The tables must be formatted precisely as outlined by the rules and examples provided. Mistakes that involve solely errant whitespace will be categorized as Presentation Error; all other errors will be reported as Wrong Answer.
题目大意：将第一个乘法按照图形输出，第二个乘数在上面横放，乘数在最右边竖放。详解如图。
只要按照红色圈圈这样就ok了，一道很麻烦的模拟题~
详见代码。
1 #include<iostream>
2 #include<stdio.h>
3 #include<string.h>
4 #define N 10000
5 using namespace std;
6 char ch[100][100];
7 int aaa[15],bbb[15],ccc[400];
8 int mul[400];
9 int main()
10 {
11 int n,m;
12 while(~scanf("%d%d",&n,&m))
13 {
14 if(n==0&&m==0) break;
15 bool flag=0;
16 int a=1,b=1,cc=10,c=1,sum=n*m;
17 memset(ccc,0,sizeof(ccc));
18 while(sum)
19 {
20 ccc[c++]=sum%cc;
21 sum/=cc;
22 //cout<<ccc[c-1]<<endl;
23 }
24 c--;
25 int kkk;
26 while(n)
27 {
28 aaa[a++]=n%cc;
29 n/=cc;
30 //cout<<aaa[a-1]<<endl;
31 }
32 a--;
33 while(m)
34 {
35 bbb[b++]=m%cc;
36 m/=cc;
37 }
38 b--;
39 kkk=a+b;
40 //for(int i=kkk;i>0;i--)
41 //cout<<ccc[i]<<endl;
42 int flagg=1,flag1=0;
43 int kk2=0;
44 for(int i=b;i>=1;i--)
45 for(int j=a;j>=1;j--)
46 mul[++kk2]=aaa[j]*bbb[i];//cout<<" "<<aaa[j]<<" "<<bbb[i]<<" "<<mul[kk2]<<endl;
47 int kk=a,kk3=1;
48 kk2=1;
49 int aa=b*4+5,bb=a*4+5;
50 for(int i=0;i<aa;i++)
51 {
52 for(int j=0;j<bb;j++)
53 {
54 if(i==0||i==aa-1)
55 {
56 if(j==0||j==bb-1)
57 printf("+");
58 else printf("-");
59 }
60 else if((i-2)%4==0&&i!=aa-2)
61 {
62 if(j==0||j==bb-1)
63 printf("|");
64 else if((j-2)%4==0)
65 printf("+");
66 else if(j==1||j==bb-2)
67 printf(" ");
68 else printf("-");
69 }
70 else if(i==1)
71 {
72 if(j==0||j==bb-1)
73 printf("|");
74 else if((j-4)%4==0)
75 printf("%d",aaa[kk--]);
76 else printf(" ");
77 }
78 else if((i-4)%4==0&&i!=aa-2)
79 {
80 if(j==0||j==bb-1)
81 printf("|");
82 else if(j==bb-2)
83 printf("%d",bbb[b-(i-4)/4]);
84 else if((j-2)%4==0)
85 printf("|");
86 else if((j-4)%4==0&&j!=bb-2&&j!=1)
87 printf("/");
88 else printf(" ");
89 }
90 else if((i-3)%4==0&&i!=aa-2)
91 {
92 if(j==0||j==bb-1)
93 printf("|");
94 else if((j-2)%4==0)
95 printf("|");
96 else if((j-3)%4==0&&j!=bb-2)
97 printf("%d",mul[kk2++]/10);
98 else if((j-5)%4==0&&j!=bb-2)
99 {
100 if(j!=1||(j==1&&flag1==1))
101 printf("/");
102 else
103 printf(" ");
104 }
105 else printf(" ");
106 }
107 else if((i-5)%4==0&&i!=aa-2)
108 {
109 //cout<<ccc[kkk]<<endl;
110 if(j==0||j==bb-1)
111 printf("|");
112 else if((j-2)%4==0)
113 printf("|");
114 else if(ccc[kkk]>=0&&j==1)
115 {
116 if(ccc[kkk]==0&&flag1==0)
117 kkk--,printf(" ");
118 else
119 printf("%d",ccc[kkk--]),flag1=1;//cout<<endl;
120 }
121 else if((j-5)%4==0&&j!=bb-2&&j!=1)
122 printf("%d",mul[kk3]%10),kk3++;
123 else if((j-3)%4==0&&j!=bb-2)
124 printf("/");
125 else printf(" ");
126 }
127 else if(i==aa-2)
128 {
129 if(j==0||j==bb-1)
130 printf("|");
131 else if(j==1&&flag==1)
132 printf("/");
133 else if((j-3)%4==0&&j!=bb-2&&ccc[kkk]>=0)
134 {
135
136 if(ccc[kkk]==0&&flag1==0)
137 kkk--,printf(" ");
138 else
139 printf("%d",ccc[kkk--]),flag1=1;
140 }
141 else if((j-5)%4==0&&j<bb-5&&flag1==1)
142 printf("/");
143 else printf(" ");
144 }
145 }
146 printf("\n");
147 }
148 }
149 return 0;
150 }
151
152 /**************************************************************
153 Problem: 1561
154 User: star_sky
155 Language: C++
156 Result: Accepted
157 Time:0 ms
158 Memory:1496 kb
159 ****************************************************************/
Problem B: Fun House
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 181 Solved: 66
[Submit][Status][Web Board]
Description
American Carnival Makers Inc. (ACM) has a long history of designing rides and attractions. One of their more popular attractions is a fun house that includes a room of mirrors. Their trademark is to set up the room so that when looking forward from the entry door, the exit door appears to be directly ahead. However, the room has double-sided mirrors placed throughout at 45 degree angles. So, the exit door can be on any of the walls of the room. The set designer always places the entry and mirrors, but can never seem to be bothered to place the exit door. One of your jobs as part of the construction crew is to determine the placement of the exit door for the room given an original design.
The final diagram for a sample room is given below. The asterisk (*) marks the entry way, lower case x's mark the walls, the mirrors are given by the forward and backward slash characters (/ and \), open spaces with no visual obstructions are marked by periods (.), and the desired placement of the exit is marked with an ampersand (&). In the input diagram, there is an 'x' in place of the '&', since the exit has not yet been located. You need to alter the input diagram by replacing the proper 'x' with an '&' to identify the exit. Note that entrances and exits can appear on any of the walls (although never a corner), and that it is physically impossible for the exit to be the same as the entrance. (You don't need to understand why this is so, although it may be fun to think about.)
xxxxxxxxxxx
x../..\...x
x..../....x
*../......x
x.........x
xxxxxx&xxxx
Input
Each room will be preceded by two integers, W and L, where 5 ≤ W ≤ 20 is the width of the room including the border walls and 5 ≤ L ≤ 20 is the length of the room including the border walls. Following the specification of W and L are L additional lines containing the room diagram, with each line having W characters from the alphabet: { * , x , . , / , \ }. The perimeter will always be comprised of walls, except for one asterisk (*) which marks the entrance; the exit is not (yet) marked. A line with two zeros indicates the end of input data.
Output
For each test case, the first line will contain the word, HOUSE, followed by a space and then an integer that identifies the given fun house sequentially. Following that should be a room diagram which includes the proper placement of the exit door, as marked by an ampersand (&).
Sample Input
11 6
xxxxxxxxxxx
x../..\...x
x..../....x
*../......x
x.........x
xxxxxxxxxxx
5 5
xxxxx
*...x
x...x
x...x
xxxxx
5 5
xxxxx
x./\x
*./.x
x..\x
xxxxx
6 6
xxx*xx
x/...x
x....x
x/./.x
x\./.x
xxxxxx
10 10
xxxxxxxxxx
x.../\...x
x........x
x........x
x.../\..\x
*...\/../x
x........x
x........x
x...\/...x
xxxxxxxxxx
0 0
Sample Output
HOUSE 1
xxxxxxxxxxx
x../..\...x
x..../....x
*../......x
x.........x
xxxxxx&xxxx
HOUSE 2
xxxxx
*...&
x...x
x...x
xxxxx
HOUSE 3
xxxxx
x./\x
*./.x
x..\&
xxxxx
HOUSE 4
xxx*xx
x/...x
x....x
x/./.&
x\./.x
xxxxxx
HOUSE 5
xxxxxxxxxx
x.../\...x
x........x
x........x
&.../\..\x
*...\/../x
x........x
x........x
x...\/...x
xxxxxxxxxx
HINT
In both Java and C++ the backslash character (\) has special meaning as an escape character within character and string literals. You must use the combination \\ to express a single backslash within a character or string literal within source code.
题目大意：主要理解"/"还有"\"这两个的表达的含义,简单的说就是初中物理学的反射，还是看图~~
从这个图片就很明显的可以看出反射之后的情况了，另一边的大同小异。*表示起点，题目是想要我们找到一个能从起点直接看到的终点（这个终点就是我们要找的），找到后将X变成&！一路看过去遇到/还有\就可以反射，转到其他的方向上面去~
详见代码。
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4
5 using namespace std;
6
7 int dir[4][2]={-1,0,1,0,0,-1,0,1};
8 int n,m,ei,ej;
9 char Map[25][25],flag1;
10
11 void Dfs(int i,int j,int ans)
12 {
13 if(flag1==1)
14 return ;
15 if(Map[i][j]=='x')
16 {
17 Map[i][j]='&';
18 flag1=1;
19 return ;
20 }
21 if(Map[i][j]=='/')
22 {
23 if(ans==0)
24 Dfs(i+dir[3][0],j+dir[3][1],3);
25 else if(ans==1)
26 Dfs(i+dir[2][0],j+dir[2][1],2);
27 else if(ans==2)
28 Dfs(i+dir[1][0],j+dir[1][1],1);
29 else if(ans==3)
30 Dfs(i+dir[0][0],j+dir[0][1],0);
31 }
32 else if(Map[i][j]=='\\')
33 {
34 if(ans==0)
35 Dfs(i+dir[2][0],j+dir[2][1],2);
36 else if(ans==1)
37 Dfs(i+dir[3][0],j+dir[3][1],3);
38 else if(ans==2)
39 Dfs(i+dir[0][0],j+dir[0][1],0);
40 else if(ans==3)
41 Dfs(i+dir[1][0],j+dir[1][1],1);
42 }
43 else
44 Dfs(i+dir[ans][0],j+dir[ans][1],ans);
45 }
46
47 int main()
48 {
49 int si,sj,flag=1;
50 while(scanf("%d%d",&n,&m),n||m)
51 {
52 for(int i=0;i<m;i++)
53 {
54 scanf("%s",Map[i]);
55 for(int j=0;j<n;j++)
56 if(Map[i][j]=='*')
57 si=i,sj=j;
58 }
59 int ans;
60 flag1=0;
61 if(si==0) ans=1;
62 else if(si==m-1) ans=0;
63 else if(sj==0) ans=3;
64 else if(sj==n-1) ans=2;
65 //cout<<ans<<" "<<dir<<endl;
66 Dfs(si,sj,ans);
67 printf("HOUSE %d\n",flag++);
68 for(int i=0;i<m;i++)
69 printf("%s\n",Map[i]);
70 }
71 return 0;
72 }
73
74 /**************************************************************
75 Problem: 1562
76 User: star_sky
77 Language: C++
78 Result: Accepted
79 Time:0 ms
80 Memory:1484 kb
81 ****************************************************************/
Problem E: Word Cloud
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 73 Solved: 37
[Submit][Status][Web Board]
Description
A word cloud (or tag cloud) is a visual representation of textual data based on a weighted metric. In the above cloud (which is based on this year's list of Mid-Central teams), the font size of each word is based on its number of occurrences in the data set. Tagg Johnson is a man obsessed with counting words that appear in online documents. On his computer, he keeps a spreadsheet of all the sites he visits, along with a list of words that appear on each site and the number of times such word appears. Tagg would like to generate word clouds based on the data he has collected.
Before describing the algorithm Tagg uses for generating clouds, we digress for a quick lesson in typography. The basic unit of measure is known as a point (typically abbreviated as pt). A font's size is described based on the vertical number of points from one line to the next, including any interline spacing. For example, with a 12pt font, the vertical space from the top of one character to the top of a character below it is 12 points. We assume that a character's height is precisely equal to the font's point size (regardless of whether the character is upper or lower case).
For this problem, we focus on a fixed-width font, such as Courier, in which each character of the alphabet is also given the same amount of width. The character width for such a font depends on the font size and the aspect ratio. For Courier, a word with t characters rendered in a font of size P has a total width of 
when measured in points. Note well the use of the ceiling operator, which converts any noninteger to the next highest integer. For example, a 5-letter word in a 20pt font would be rendered with a height of 20 points and a width equal to 
points.
Now we can describe Tagg's algorithm for creating a word cloud. He pre-sorts his word list into alphabetical order and removes words that do not occur at least five times. For each word w, he computes a point size based on the formula 
, where cw is the number of occurrences of the word, and cmax is the number of occurrences of the most frequent word in the data set. Note that by this formula, every word will be rendered with anywhere from a 9pt font to a 48pt font. He then places the words in rows, with a 10pt horizontal space between adjacent words, placing as many words as fit in the row, subject to a maximum width W for his entire cloud. The height of a given row is equal to the maximum font size of any word rendered in that row.
As a tangible example, consider the following data set and word cloud.
Word Count
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
In this example, apple is rendered with 23pt font using width 65pt, banana is rendered with 11pt font using width 38pt, and grape is rendered with 48pt font and width 135pt. If the overall word cloud is constrained to have width at most 260, those three words fit in a row and the overall height of that row is 48pt (due to grape). On the second row kiwi is rendered with height 43pt and width 97pt, and orange is rendered with height 28pt and width 95pt. A third row has strawberry with height 23pt and width 130pt. The overall height of this word cloud is 114pt.
Input
Each data set begins with a line containing two integers: W and N. The value W denotes the maximum width of the cloud; W ≤ 5000 will be at least as wide as any word at its desired font size. The value 1 ≤ N ≤ 100 denotes the number of words that appear in the cloud. Following the first line are N additional lines, each having a string S that is the word (with no whitespace), and an integer C that is a count of the number of occurrences of that word in the original data set, with 5 ≤ C ≤ 1000. Words will be given in the same order that they are to be displayed within the cloud.
Output
For each data set, output the word CLOUD followed by a space, a serial number indicating the data set, a colon, another space, and the integer height of the cloud, measured in font points.
Sample Input
260 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
250 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
610 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
0 0
Sample Output
CLOUD 1: 114
CLOUD 2: 99
CLOUD 3: 48
题目大意：把题目输入的每一个都看成一个小盒子，最外层的是一个大盒子。按照所给的顺序存放在一个盒子中，开始输入的两个w和n分别表示盒子的最大的宽度，以及有多少个盒子。然后输入每个小盒子和分别出现的次数。
说下这两个公式，第一个这个是求p（高度），cw表示的是每个小盒子的出现次数，cmax表示出现最多的次数。
第二个这个是求宽度的，t表示的是apple的字节长度，p就是上面求得那个。
通过这两个公式就可以求的每个小盒子的高度和宽度。
特别注意：1、要按照所给的顺序放，放到这一行不能放了在换行。
2、with a 10pt horizontal space between adjacent words.特别注意这句话，好久才发现的，这个是说每个小盒子间要空10，所以不要忘记加这个10，但是第一个千万不要加哦~
详见代码。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5
6 using namespace std;
7
8 char ch[11000];
9
10 int main ()
11 {
12 int w,n,t[210],p[210],num[210],time[210],flag=1;
13 while (scanf("%d%d",&w,&n),w||n)
14 {
15 int Max=0;
16 for (int i=0; i<n; i++)
17 {
18 scanf("%s%d",ch,&time[i]);
19 //memset(vis,0,sizeof(vis));
20 t[i]=strlen(ch);
21 if (Max<time[i])
22 Max=time[i];
23 }
24 for (int i=0; i<n; i++)
25 {
26 p[i]=8+ceil(40*((time[i]-4)*(1.0))/(Max-4));
27 num[i]=ceil(((9*1.0)/16)*p[i]*t[i]);
28 //cout<<num[i]<<" "<<p[i]<<" "<<(time[i]-4)*(1.0)/(Max-4)<<endl;
29 }
30 int i=0,sum=0;
31 while(i<n)
32 {
33 int s=0;
34 int maxx=0;
35 while(s<=w)
36 {
37 //cout<<s<<endl;
38 if(s+num[i]>w||i==n)
39 break;
40 s+=num[i]+10;
41 maxx=max(maxx,p[i]);
42 i++;
43 }
44 sum+=maxx;
45 }
46 printf("CLOUD %d: %d\n",flag++,sum);
47 }
48 return 0;
49 }
50
51 /**************************************************************
52 Problem: 1565
53 User: star_sky
54 Language: C++
55 Result: Accepted
56 Time:0 ms
57 Memory:1496 kb
58 ****************************************************************/
Problem G: Reverse Rot
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 150 Solved: 82
[Submit][Status][Web Board]
Description
A very simplistic scheme, which was used at one time to encode information, is to rotate the characters within an alphabet and rewrite them. ROT13 is the variant in which the characters A-Z are rotated 13 places, and it was a commonly used insecure scheme that attempted to "hide" data in many applications from the late 1990's and into the early 2000's.
It has been decided by Insecure Inc. to develop a product that "improves" upon this scheme by first reversing the entire string and then rotating it. As an example, if we apply this scheme to string ABCD with a reversal and rotation of 1, after the reversal we would have DCBA and then after rotating that by 1 position we have the result EDCB.
Your task is to implement this encoding scheme for strings that contain only capital letters, underscores, and periods. Rotations are to be performed using the alphabet order:
ABCDEFGHIJKLMNOPQRSTUVWXYZ_.
Note that underscore follows Z, and the period follows the underscore. Thus a forward rotation of 1 means 'A' is shifted to 'B', that is, 'A'→'B', 'B'→'C', ..., 'Z'→'_', '_'→'.', and '.'→'A'. Likewise a rotation of 3 means 'A'→'D', 'B'→'E', ..., '.'→'C'.
Input
Each input line will consist of an integer N, followed by a string. N is the amount of forward rotation, such that 1 ≤ N ≤ 27. The string is the message to be encrypted, and will consist of 1 to 40 characters, using only capital letters, underscores, and periods. The end of the input will be denoted by a final line with only the number 0.
Output
For each test case, display the "encrypted" message that results after being reversed and then shifted.
Sample Input
1 ABCD
3 YO_THERE.
1 .DOT
14 ROAD
9 SHIFTING_AND_ROTATING_IS_NOT_ENCRYPTING
2 STRING_TO_BE_CONVERTED
1 SNQZDRQDUDQ
0
Sample Output
EDCB
CHUHKWBR.
UPEA
ROAD
PWRAYF_LWNHAXWH.RHPWRAJAX_HMWJHPWRAORQ.
FGVTGXPQEAGDAQVAIPKTVU
REVERSE_ROT
题目大意：直接说下输入吧，输入的第一个数就是表示一次移动几格，第二个就是加了密的密码。要按照ABCDEFGHIJKLMNOPQRSTUVWXYZ_.进行译码，但是注意要从加密的密码尾部开始译码。最后输出就ok了，哇哈哈~注意一种特殊情况。
给一组测试数据
2 WFDFG..__..__ 这个就是考虑，数字和字符之间会不会有多个空格，解决这个问题就没大问题了
详见代码。
1 #include<iostream>
2 #include<stdio.h>
3 #include<string.h>
4 #define N 10000
5 using namespace std;
6 int main()
7 {
8 char ch[50]={"ABCDEFGHIJKLMNOPQRSTUVWXYZ_.\0"};
9 int n;
10 char a[100];
11 while(~scanf("%d%s",&n,a))
12 {
13 if(n==0) break;
14 //cout<<a<<endl;
15 int i=strlen(a),k;
16 for(int j=0;j<i;j++)
17 {
18 if(a[j]=='_')
19 {
20 k=26;
21 //cout<<ch[k]<<endl;
22 }
23 else if(a[j]=='.')
24 {
25 k=27;
26 //cout<<ch[k]<<endl;
27 }
28 else
29 {
30 k=a[j]-'A';
31 }
32 k+=n;
33 k%=28;
34 a[j]=ch[k];
35 }
36 for(int j=i-1;j>=0;j--)
37 printf("%c",a[j]);
38 printf("\n");
39 }
40 return 0;
41 }
42
43 /**************************************************************
44 Problem: 1567
45 User: star_sky
46 Language: C++
47 Result: Accepted
48 Time:0 ms
49 Memory:1484 kb
50 ****************************************************************/
Problem I: Wet Tiles
Time Limit: 60 Sec Memory Limit: 512 MB
Submit: 73 Solved: 34
[Submit][Status][Web Board]
Description
Alice owns a construction company in the town of Norainia, famous for its unusually dry weather. In fact, it only rains a few days per year there. Because of this phenomenon, many residents of Norainia neglect to do roof repairs until leaks occur and ruin their floors. Every year, Alice receives a deluge of calls from residents who need the leaks fixed and floor tiles replaced. While exquisite in appearance, Norainia floor tiles are not very water resistant; once a tile becomes wet, it is ruined and must be replaced. This year, Alice plans to handle the rainy days more efficiently than in past years. She will hire extra contractors to dispatch as soon as the calls come in, so hopefully all leaks can be repaired as soon as possible. For each house call, Alice needs a program to help her determine how many replacement tiles a contractor team will need to bring to complete the job.
For a given house, square floor tiles are arranged in a rectangular grid. Leaks originate from one or more known source locations above specific floor tiles. After the first minute, the tiles immediately below the leaks are ruined. After the second minute, water will have spread to any tile that shares an edge with a previously wet tile. This pattern of spreading water continues for each additional minute. However, the walls of a house restrict the water; if a damaged area hits a wall, the water does not penetrate the wall. We assume there are always four outer walls surrounding the entire house. A house may also have a number of additional "inner" walls; each inner wall is comprised of a connected linear sequence of locations (which may or may not be connected to the outer walls or to each other).
As an example, Figure 1 shows water damage (in gray) that would result from three initial leaks (each marked with a white letter 'L') after each of the first five minutes of time. Tiles labeled '2' become wet during the second minute, tiles labeled '3' become wet during the third minute, and so forth. The black areas designate inner walls that restrict the flow of water. Note that after 5 minutes, a total of 75 tiles have been damaged and will need to be replaced. Figures 2 through 4 show other houses that correspond to the example inputs for this problem.
75 wet tiles
17 wet tiles
4 wet tiles
94 wet tiles
Input
Each house is described beginning with a line having five integral parameters: X Y T L W. Parameters X and Y designate the dimensions of the rectangular grid, with 1 ≤ X ≤ 1000 and 1 ≤ Y ≤ 1000. The coordinate system is one-indexed, as shown in the earlier figures. Parameter T designates the number of minutes that pass before a team of contractors arrives at a house and stops the leaks, with 1 ≤ T ≤ 200000. The parameter L designates the number of leaks, with 1 ≤ L ≤ 100. Parameter W designates the number of inner walls in the house, 0 ≤ W ≤ 100.
The following 2L integers in the data set, on one or more lines, are distinct (x y) pairs that designate the locations of the L distinct leaks, such that 1 ≤ x ≤ X and 1 ≤ y ≤ Y.
If W > 0, there will be 4W additional integers, on one or more lines, that describe the locations of the walls. For each such wall the four parameters (x1,y1), (x2,y2) describe the locations of two ends of the wall. Each wall replaces a linear sequence of adjoining tiles and is either axis-aligned or intersects both axes at a 45 degree angle. Diagonal walls are modeled as a sequence of cells that would just be touching corner to corner. If the two endpoints of a wall are the same, the wall just occupies the single cell at that location. Walls may intersect with each other, but no leak is over a wall.
There will be one or more houses in the data file and a line with a single integer -1 designates the end of the data set.
Output
For each house, display the total number of tiles that are wet after T minutes.
Sample Input
12 12 5 3 5
2 11 3 3 9 5
1 9 6 9 1 7 4 4 7 1 7 4
10 9 10 12 11 4 12 4
9 7 8 1 3
4 3
2 2 6 6 6 2 2 6 8 2 8 2
6 7 50 1 3
3 4
2 2 2 6 3 6 5 4 5 4 3 2
12 12 5 3 0
2 11 3 3 9 5
-1
Sample Output
75
17
4
94
最后一题，在还有半个小时的时候才看的，题目太长猜了一下，结果理解了题意0.0
题目大意:给一个x*y的矩阵，给出几个漏水的地点，还会给出墙的起始坐标。漏水点水会蔓延，到墙的位置过不去，求有多少瓷砖被淹。用广搜就可以，起点已经给了、
详见代码。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<queue>
5 using namespace std;
6 struct ssss
7 {
8 int x,y,t;
9 } s,ss;
10 bool Map[1010][1010];
11 int xy[4][2]= {0,1,0,-1,1,0,-1,0};
12 int main()
13 {
14 int n,m,t,l,w;
15 while(~scanf("%d",&n))
16 {
17 if(n==-1) break;
18 scanf("%d%d%d%d",&m,&t,&l,&w);
19 queue<ssss>q;
20 memset(Map,0,sizeof(Map));
21 int sum=l;
22 while(l--)
23 {
24 scanf("%d%d",&s.x,&s.y);
25 s.y=m-s.y+1;
26 s.t=1;
27 Map[s.x][s.y]=1;
28 q.push(s);
29 }
30 while(w--)
31 {
32 //cout<<w<<endl;
33 int a,b,c,d;
34 scanf("%d%d%d%d",&a,&b,&c,&d);
35 b=m-b+1;
36 d=m-d+1;
37 int x=d-b,y=c-a;
38 if(y==0)
39 {
40 if(x>0)
41 {
42 while(b<=d)
43 {
44 Map[a][b]=1,b++;
45 }
46 }
47 else
48 while(b>=d)
49 {
50 Map[a][b]=1,b--;
51 }
52 }
53 else if(y<0)
54 {
55 if(x>0)
56 while(a>=c)
57 {
58 Map[a][b]=1,b++,a--;
59 }
60 else if(x<0)
61 while(a>=c)Map[a][b]=1,b--,a--;
62 else while(a>=c)Map[a][b]=1,a--;
63 }
64 else
65 {
66 if(x>0)
67 while(a<=c)
68 Map[a][b]=1,b++,a++;
69 else if(x<0)
70 {
71 while(a<=c) Map[a][b]=1,b--,a++;
72 }
73 else while(a<=c)Map[a][b]=1,a++;
74
75 }
76 }
77 while(q.size())
78 {
79 s=q.front();
80 q.pop();
81 for(int i=0; i<4; i++)
82 {
83 ss.x=s.x+xy[i][0];
84 ss.y=s.y+xy[i][1];
85 ss.t=s.t+1;
86 if(ss.x>=1&&ss.x<=n&&ss.y>=1&&ss.y<=m&&ss.t<=t&&Map[ss.x][ss.y]==0)
87 {
88 sum++;
89 Map[ss.x][ss.y]=1;
90 q.push(ss);
91 }
92 }
93 }
94 printf("%d\n",sum);
95 }
96 }
97
98 /**************************************************************
99 Problem: 1569
100 User: star_sky
101 Language: C++
102 Result: Accepted
103 Time:944 ms
104 Memory:2488 kb
105 ****************************************************************/
