pku2894-2903 Tehran 2005 比赛总结

Solved	ID	Title	Ratio(AC/att)
Yes	Problem A	Ancient Keyboard	100.0%(1/1)	Submit
Yes	Problem B	Best SMS to Type	50.0%(1/2)	Submit
Yes	Problem C	Changing Phone Numbers	100.0%(1/1)	Submit
Yes	Problem D	Dramatic Multiplications	33.33%(1/3)	Submit
Yes	Problem E	Entertainment	100.0%(1/1)	Submit
Yes	Problem F	Fortune at El Dorado	100.0%(1/1)	Submit
Yes	Problem G	Griddy Hobby	100.0%(1/1)	Submit
	Problem H	Hotel	0.0%(0/0)	Submit
	Problem I	Intercepting Missiles	0.0%(0/0)	Submit
Yes	Problem J	Joy of Mobile Routing	100.0%(1/1)	Submit

这场比赛题目难度适中，有2题没有做出来。。
pku2894 Ancient Keyboard
直接开个数组来模拟就可以了，数据量再大的话就线段树吧- -
代码：

# include < iostream >
2

# include < cstring >
3

using namespace std;
4

int main()
5

{
6

int test;
7

cin>>test;
8

while(test--)
9

{
10

int n;
11

cin>>n;
12

char l;
13

int s,e,c[1001],max=-1,min=0xfffffff;
14

memset(c,0,sizeof(c));
15

while(n--)
16

{
17

cin>>l>>s>>e;
18

for(int i=s;i<e;i++)
19

c[i]++;
20

if(e>max) max=e;
21

if(s<min) min=s;
22

}
23

for(int i=min;i<=max;i++)
24

if(c[i])
25

cout<<(char)(c[i]+64);
26

cout<<endl;
27

}
28

return 0;
29

}

pku2896 Best SMS to Type
同样是个没有难度的模拟，细心点就可以了~

# include < iostream >
2

using namespace std;
3

inline bool equal( int a, int b)
4

{
5

if(a<15&&b<15&&a/3==b/3) return true;
6

else if(a<19&&b<19&&a>=15&&b>=15) return true;
7

else if(a<22&&b<22&&a>=19&&b>=19) return true;
8

else if(a>=22&&b>=22) return true;
9

else return false;
10

}
11

int main()
12

{
13

int test,map[255];
14

map['A']=map['D']=map['G']=map['J']=map['M']=map['P']=map['T']=map['W']=1;
15

map['B']=map['E']=map['H']=map['K']=map['N']=map['Q']=map['U']=map['X']=2;
16

map['C']=map['F']=map['I']=map['L']=map['O']=map['R']=map['V']=map['Y']=3;
17

map['S']=map['Z']=4;
18

map[' ']=1;
19

cin>>test;
20

while(test--)
21

{
22

int t1,t2,total=0;
23

char str[1024];
24

cin>>t1>>t2;
25

cin.get();
26

cin.getline(str,1024);
27

total+=t1*map[str[0]];
28

for(int i=1;str[i]!='\0';i++)
29

if(str[i]!=' '&&str[i-1]!=' '&&equal(str[i]-65,str[i-1]-65))
30

total+=map[str[i]]*t1+t2;
31

else total+=map[str[i]]*t1;
32

cout<<total<<endl;
33

}
34

return 0;
35

}

pku2896 Changing Phone Numbers
各种数据结构的应用，先按时间排序，然后借助于堆、字典树之类的东西设计一个离线算法应该不难。记得看清题目： [s+1,e]区间内的规则适用于一个询问。
代码：

import java.util. * ;
2

import java.io. * ;
3

class dic
4

{
5

class node
6

{
7

node nxt[]=new node[10];
8

String str;
9

boolean end=false;
10

node()
11

{
12

Arrays.fill(nxt,null);
13

}
14

}
15

node head=new node();
16

void insert(String num,String str)
17

{
18

node p=head;
19

for(int i=0;i<num.length();i++)
20

{
21

if(p.nxt[num.charAt(i)-'0']==null)
22

p.nxt[num.charAt(i)-'0']=new node();
23

p=p.nxt[num.charAt(i)-'0'];
24

}
25

p.end=true;
26

p.str=str;
27

}
28

void erase(String num)
29

{
30

node p=head;
31

for(int i=0;i<num.length();i++)
32

p=p.nxt[num.charAt(i)-'0'];
33

p.end=false;
34

}
35

String get(String num)
36

{
37

node p=head;
38

for(int i=0;i<num.length();i++)
39

{
40

if(p.end) return p.str;
41

p=p.nxt[num.charAt(i)-'0'];
42

}
43

return p.str;
44

}
45

}
46

public class Main

{
47

/** *//**
49

* @param argsi
50

*/
51

static class ins implements Comparable<ins>
53

{
54

int year,type;
55

String name,op;
56

public int compareTo(ins pos)
57

{
58

return year-pos.year;
59

}
60

ins(int year,int type,String name,String op)
61

{
62

this.year=year;
63

this.type=type;
64

this.name=name;
65

this.op=op;
66

}
67

}
68

static class node
69

{
70

int s,e,id;
71

String code;
72

node(int id,int s,int e,String code)
73

{
74

this.s=s;
75

this.e=e;
76

this.code=code;
77

this.id=id;
78

}
79

}
80

static class local
81

{
82

String num;
83

TreeSet<node> list=new TreeSet<node>(new cmp_e());
84

local(String str)
85

{
86

num=str;
87

}
88

void ChangeCode(String code)
89

{
90

String name=antirefer.get(num);
91

antirefer.erase(num);
92

num=code;
93

antirefer.insert(num, name);
94

}
95

void add(node pos)
96

{
97

pos.code=pos.code.substring(num.length());
98

list.add(pos);
99

}
100

void repeat(int pos)
101

{
102

for(node p:list)
103

p.code=p.code.substring(0,pos)+p.code.substring(pos-1);
104

}
105

void swap(int pos)
106

{
107

for(node p:list)
108

p.code=p.code.substring(0,pos-1)+p.code.charAt(pos)+p.code.charAt(pos-1)+p.code.substring(pos+1);
109

}
110

void move(int year)
111

{
112

while(!list.isEmpty()&&list.first().e<year)
113

{
114

list.first().code=num+list.first().code;
115

ans.add(list.pollFirst());
116

}
117

118

119

}
120

}
121

static class cmp_s implements Comparator<node>
122

{
123

public int compare(node a,node b)
124

{
125

return a.s-b.s;
126

}
127

}
128

static class cmp_e implements Comparator<node>
129

{
130

public int compare(node a,node b)
131

{
132

if(a.e!=b.e) return a.e-b.e;
133

else return a.id-b.id;
134

}
135

}
136

static class cmp_id implements Comparator<node>
137

{
138

public int compare(node a,node b)
139

{
140

return a.id-b.id;
141

}
142

}
143

static dic antirefer=new dic();
144

static ArrayList<node> ans=new ArrayList<node>();
145

public static void main(String[] args) throws IOException

{
146

Scanner in=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
147

HashMap<String,local> refer=new HashMap<String,local>();
148

ArrayList<ins> list=new ArrayList<ins>();
149

ArrayList<node> ques=new ArrayList<node>();
150

int n=in.nextInt();
151

while((n--)!=0)
152

{
153

String code=in.next(),name=in.next();
154

refer.put(name, new local(code));
155

antirefer.insert(code, name);
156

}
157

n=in.nextInt();
158

while((n--)!=0)
159

list.add(new ins(in.nextInt(),in.nextInt(),in.next(),in.next()));
160

int tc=0;
161

while(true)
162

{
163

int s=in.nextInt(),e=in.nextInt();
164

String code=in.next();
165

if(s==0&&e==0&&code.equals("0")) break;
166

ques.add(new node(tc++,s,e,code));
167

}
168

Collections.sort(list);
169

Collections.sort(ques,new cmp_s());
170

int p=0;
171

for(ins i:list)
172

{
173

while(p<ques.size()&&ques.get(p).s<i.year)
174

{
175

if(antirefer.get(ques.get(p).code)==null)
176

{
177

System.out.println("wa");
178

}
179

refer.get(antirefer.get(ques.get(p).code)).add(ques.get(p));
180

p++;
181

}
182

refer.get(i.name).move(i.year);
183

switch(i.type)
184

{
185

case 1:
186

refer.get(i.name).repeat(Integer.parseInt(i.op));
187

break;
188

case 2:
189

refer.get(i.name).swap(Integer.parseInt(i.op));
190

break;
191

case 3:
192

refer.get(i.name).ChangeCode(i.op);
193

break;
194

};
195

196

}
197

for(;p<ques.size();p++)
198

ans.add(ques.get(p));
199

for(local i:refer.values())
200

i.move(Integer.MAX_VALUE);
201

Collections.sort(ans,new cmp_id());
202

for(node i:ans)
203

System.out.println(i.code);
204

205

}
206

207

}

pku2897 Dramatic Multiplications
一道数学题，MS现在权位展开的题目不少- -。这题可以从低位向高位逐位确定。
   AP
*   K
-------
   PA
什么情况无解？出现了循环~。注意！第一位不能为0
代码：

# include < iostream >
2

# include < stack >
3

# include < cstring >
4

using namespace std;
5

bool used[ 10 ][ 10 ];
6

int main()
8

{
9

int test;
10

cin>>test;
11

while(test--)
12

{
13

int n,k,last,add;
14

bool flag=false;
15

cin>>n>>k;
16

stack<int> ans;
17

memset(used,false,sizeof(used));
18

last=k,add=0;
19

while(!used[last][add])
20

{
21

used[last][add]=true;
22

int nxt=last*n+add;
23

last=nxt%10;
24

add=nxt/10;
25

if(!add&&last==k&&(ans.empty()||ans.top()))
26

{
27

for(;!ans.empty();ans.pop())
28

cout<<ans.top();
29

cout<<k<<endl;
30

flag=true;
31

break;
32

}
33

else ans.push(last);
34

}
35

if(!flag) cout<<0<<endl;
36

}
37

return 0;
38

}

pku2898 Entertainment
又是一个模拟的题目，记得POJ上有题叫什么game的和这个一样。用数组模拟就可以了。好久不写BFS，忘了一点，BFS判重的时候是在进队之前~，还有就是，字符输入输出尽量用gets,puts,getchar,putchar

代码：

# include <cstdio>
2

# include <cstring>
3

# include <cstdlib>
4

# define max(a,b) ((a)>(b)?(a):(b))
5

# define min(a,b) ((a)<(b)?(a):(b))
6

# define legal(r,c) ((r)>=start&&(r)<n&&(c)>=0&&(c)<m&&!inqueue[r][c])
7

using namespace std;
8

const int N=1024;
9

char map[N][N];
10

int n,m,k,start,q[N*N][2],s,e;
11

bool inqueue[N][N];
12

void bfs(int tr,int tc)
13

{
14

s=e=-1;
16

q[++e][0]=tr;
17

q[e][1]=tc;
18

memset(inqueue,false,sizeof(inqueue));
19

while(s!=e)
20

{
21

s++;
22

tr=q[s][0];
23

tc=q[s][1];
24

inqueue[tr][tc]=true;
25

if(legal(tr-1,tc)&&map[tr-1][tc]==map[tr][tc])
26

{
27

q[++e][0]=tr-1;
28

q[e][1]=tc;
29

}
30

if(legal(tr+1,tc)&&map[tr+1][tc]==map[tr][tc])
31

{
32

q[++e][0]=tr+1;
33

q[e][1]=tc;
34

}
35

if(legal(tr,tc-1)&&map[tr][tc-1]==map[tr][tc])
36

{
37

q[++e][0]=tr;
38

q[e][1]=tc-1;
39

}
40

if(legal(tr,tc+1)&&map[tr][tc+1]==map[tr][tc])
41

{
42

q[++e][0]=tr;
43

q[e][1]=tc+1;
44

}
45

map[tr][tc]=' ';
46

}
47

}
48

void MoveLeft()
49

{
50

int maxnum=-1;
51

for(int i=start;i<n;i++)
52

{
53

int total=0;
54

for(int j=m-1;j>=0;j--)
55

if(map[i][j]==' ')
56

{
57

for(int k=j+1;k<m;k++)
58

map[i][k-1]=map[i][k],map[i][k]=' ';
59

total++;
60

}
61

maxnum=max(maxnum,m-total);
62

}
63

m=maxnum;
64

}
65

void MoveDown()
66

{
67

int minnum=0xfffffff;
68

for(int j=0;j<m;j++)
69

{
70

int total=0;
71

for(int i=start;i<n;i++)
72

if(map[i][j]==' ')
73

{
74

for(int k=i-1;k>=start;k--)
75

map[k+1][j]=map[k][j],map[k][j]=' ';
76

total++;
77

}
78

minnum=min(minnum,start+total);
79

}
80

start=minnum;
81

}
82

void print()
83

{
84

for(int i=start;i<n;i++)
85

{
86

for(int j=0;j<m;j++)
87

putchar(map[i][j]);
88

putchar('\n');
89

}
90

// printf("\n");
91

}
92

int main()
93

{
94

//freopen("e.in","r",stdin);
95

//freopen("ans.txt","w",stdout);
96

int test=0;
97

while(gets(map[0]))
98

{
99

m=strlen(map[0]);
100

for(n=1;;n++)
101

{
102

gets(map[n]);
103

if(map[n][0]>='0'&&map[n][0]<='9')
104

{
105

k=atoi(map[n]);
106

break;
107

}
108

}
109

110

start=0;
111

for(int i=0;i<k;i++)
112

{
113

int tr,tc;
114

scanf("%d%d",&tr,&tc);
115

tr--,tc--;
116

117

bfs(tr+start,tc);
118

// print();
119

// system("pause");
120

MoveLeft();
121

// print();
122

// system("pause");
123

MoveDown();
124

// print();
125

// system("pause");
126

127

}
128

printf("Test case #%d:\n",++test);
129

print();
130

if(k) getchar();
131

//memset(map,0,sizeof(map));
132

}
133

}

pku2899 Fortune at El Dorado
经典的二维动态统计：扫描线+树状数组。
这题有点新意的地方就是要枚举扫描线宽度，然后以前动态维护一个树状数组的算法有点不适用了，所以就开了n个树状数组，然后相减即可~复杂度仍然为o(nlogn)，BS这题给出的数据范围，点总数只有100个却写1000个。。1000个点估计没有什么好算法了吧。。
代码：

# include <cstdio>
2

# include <utility>
3

# include <cstring>
4

# include <functional>
5

# include <algorithm>
6

using namespace std;
7

# define lowbit(bit) ((bit)&-(bit))
8

# define N 1001
9

struct node
10

{
11

int arr[N];
12

int x;
13

void add(int pos)
14

{
15

while(pos<N)
16

arr[pos]++,pos+=lowbit(pos);
17

}
18

}refer[N];
20

int n,aera,c;
21

pair<int,int> data[N];
22

bool cmp(const pair<int,int> &a,const pair<int,int> &b)
23

{
24

return a.first<b.first;
25

}
26

int sum(int pos,int s,int e)
27

{
28

int res=0;
29

while(pos>0)
30

res+=refer[e].arr[pos]-refer[s-1].arr[pos],pos-=lowbit(pos);
31

return res;
32

}
33

int get(int rank,int s,int e)
34

{
35

int begin=1,end=N-1;
36

while(begin<=end)
37

{
38

int mid=(begin+end)/2;
39

if(sum(mid,s,e)>=rank) end=mid-1;
40

else begin=mid+1;
41

}
42

return end+1;
43

}
44

int main()
45

{
46

//freopen("ans.txt","w",stdout);
47

//freopen("f.in","r",stdin);
48

int test;
49

scanf("%d",&test);
50

while(test--)
51

{
52

scanf("%d%d",&n,&aera);
53

for(int i=0;i<n;i++)
54

scanf("%d%d",&data[i].first,&data[i].second);
55

sort(data,data+n,cmp);
56

c=2;
57

memset(refer[1].arr,0,sizeof(refer[1].arr));
58

memset(refer[0].arr,0,sizeof(refer[0].arr));
59

refer[1].x=data[0].first;
60

refer[1].add(data[0].second);
61

for(int i=1;i<n;i++)
62

if(data[i].first==data[i-1].first)
63

refer[c-1].add(data[i].second);
64

else
65

{
66

memcpy(refer[c].arr,refer[c-1].arr,sizeof(refer[c].arr));
67

refer[c].x=data[i].first;
68

refer[c++].add(data[i].second);
69

}
70

int ans=-1;
71

for(int i=1;i<c;i++)
72

for(int j=i;j<c;j++)
73

{
74

int rank=sum(N-1,i,j);
75

int len=refer[j].x-refer[i].x?aera/(refer[j].x-refer[i].x):aera;
76

while(rank)
77

{
78

int p=get(rank,i,j);
79

ans=max(ans,sum(p,i,j)-sum(p-len-1,i,j));
80

rank--;
81

}
82

}
83

printf("%d\n",ans);
84

}
85

return 0;
86

}

pku2900 Griddy Hobby
有趣的题目，一条光线从一点45度角射出，不断反弹，直到循环或者射出为止。求包含的最小长方形数目。我用的方法很笨。。将所有的线段模拟出来后然后整个坐标系旋转45度，用上题的动态统计的方法做的。。MS可以直接根据交点数来算。。不过我的那种方法更通用
模拟的时候考虑周全一点，四种向量，8种情况（每一个向量落在不同的边上时旋转地方向不同）

代码：

# include <set>
2

# include <iostream>
3

# include <algorithm>
4

# include <string>
5

# include <vector>
6

# include <utility>
7

# include <functional>
8

# include <cstring>
9

# include <queue>
10

# define lowbit(bit) ((bit)&-(bit))
11

using namespace std;
12

struct node
13

{
14

int r1,c1,r2,c2;
15

int x1,y1,x2,y2;
16

node(int r1,int c1,int r2,int c2)
17

{
18

this->r1=r1;this->c1=c1;
19

this->r2=r2;this->c2=c2;
20

}
21

void trans()
22

{
23

y1=-(r1-1);
24

y2=-(r2-1);
25

x1=c1-1;
26

x2=c2-1;
27

int tx,ty;
28

tx=x1-y1,ty=x1+y1;
29

x1=tx,y1=ty;
30

tx=x2-y2,ty=x2+y2;
31

x2=tx,y2=ty;
32

if(x1==x2&&y1>y2)
33

swap(y1,y2);
34

if(y1==y2&&x1>x2)
35

swap(x1,x2);
36

}
38

};
39

bool cmp_x1(const node &a,const node &b)
40

{
41

return a.x1<b.x1;
42

}
43

struct cmp_x2
44

{
45

bool operator()(const node &a,const node &b) const
46

{
47

return a.x2>b.x2;
48

}
49

};
50

vector<node> p;
51

bool used[1001][1001];
52

vector<node> x;
53

vector<node> y;
54

priority_queue<node,vector<node>,cmp_x2> q;
55

int tree[5000];
56

int sum(int pos)
57

{
58

int res=0;
59

while(pos>0)
60

res+=tree[pos],pos-=lowbit(pos);
61

return res;
62

}
63

void add(int pos,int one)
64

{
65

while(pos<=4000)
66

tree[pos]+=one,pos+=lowbit(pos);
67

}
68

int main()
69

{
70

int test;
71

cin>>test;
72

while(test--)
73

{
74

int r,c,n,m,dir;
75

string tdir;
76

cin>>n>>m>>r>>c>>tdir;
77

if(tdir=="UR") dir=0;
78

else if(tdir=="DR") dir=1;
79

else if(tdir=="DL") dir=2;
80

else dir=3;
81

p.clear();
82

x.clear();
83

y.clear();
84

memset(used,false,sizeof(used));
85

do
86

{
87

used[r][c]=true;
88

int tr,tc;
89

switch(dir)
90

{
91

case 0:
92

tr=r-min(m-c,r-1);
93

tc=c+min(m-c,r-1);
94

if(tr==1) dir=1;
95

else dir=3;
96

break;
97

case 1:
98

tr=r+min(m-c,n-r);
99

tc=c+min(m-c,n-r);
100

if(tc==m) dir=2;
101

else dir=0;
102

break;
103

case 2:
104

tr=r+min(n-r,c-1);
105

tc=c-min(n-r,c-1);
106

if(tr==n) dir=3;
107

else dir=1;
108

break;
109

case 3:
110

tr=r-min(r-1,c-1);
111

tc=c-min(r-1,c-1);
112

if(tc==1) dir=0;
113

else dir=2;
114

break;
115

};
116

p.push_back(node(r,c,tr,tc));
117

r=tr,c=tc;
118

}while(!(r==1&&c==1||r==n&&c==m||r==1&&c==m||r==n&&c==1||used[r][c]));
119

for(int i=0;i<p.size();i++)
120

{
121

p[i].trans();
122

if(p[i].x1==p[i].x2) x.push_back(p[i]);
123

else y.push_back(p[i]);
124

}
125

sort(x.begin(),x.end(),cmp_x1);
126

sort(y.begin(),y.end(),cmp_x1);
127

int j=0;
128

while(!q.empty()) q.pop();
129

memset(tree,0,sizeof(tree));
130

int res=0;
131

int size=x.size();
132

for(int i=0;i<size-1;i++)
133

{
134

while(j<y.size()&&y[j].x1<=x[i].x1)
135

{
136

add(y[j].y1+2000,1);
137

q.push(y[j++]);
138

}
139

while(!q.empty()&&q.top().x2<x[i+1].x1)
140

{
141

add(q.top().y1+2000,-1);
142

q.pop();
143

}
144

res+=sum(min(x[i].y2,x[i+1].y2)+2000)-sum(max(x[i].y1,x[i+1].y1)-1+2000)>=2?sum(min(x[i].y2,x[i+1].y2)+2000)-sum(max(x[i].y1,x[i+1].y1)-1+2000)-1:0;
145

}
146

cout<<res<<endl;
147

}
148

return 0;
149

}

pku2903 Joy of Mobile Routing
一道三维向量的题目。首先枚举岔口和发射站的工作省不了的，这里n²m²的复杂度，然后就是判断发射站能不能发射到岔口。
这里MS要详细讨论4种情况：岔口在发射站的左上、右上、左下、右下四种情况，然后上取整和下取整注意下，行列分别验证~总复杂度n³m²，还有向量的方向不能反，必须是从发射站到岔口，应为反过来会有一种很难处理的东西，就是射线根本过不了第一个障碍物，但是到达障碍物对面的时候是比它高的，这样就造成了误判。无权图最短路径不用说了吧，BFS

# include < cstdio >
2

using namespace std;
3

# include < algorithm >
4

# include < cmath >
5

# include < queue >
6

# include < cstring >
7

# include < utility >
8

# include < functional >
9

# define N 55
10

# define eps 1e - 8
11

# define equ(a,b) (fabs((a) - (b)) < eps)
12

# define gt(a,b) ( ! equ(a,b) && (a) > (b))
13

int h[N][N],len[N][N];
14

bool used[N][N];
15

int n,m,num,sr,sc,er,ec;
16

struct node
17

{
18

int r,c,h;
19

bool operator<(const node &pos) const
20

{
21

return h>pos.h;
22

}
23

} sta[ 105 ];
24

int bfs()
25

{
26

used[er][ec]=true;
27

if(!used[sr][sc]||!used[er][ec]) return -1;
28

queue<pair<int,int> > q;
29

memset(len,-1,sizeof(len));
30

len[sr][sc]=0;
31

q.push(make_pair<int,int>(sr,sc));
32

while(!q.empty())
33

{
34

pair<int,int> top=q.front();
35

q.pop();
36

if(top.first-1>=0&&used[top.first-1][top.second]&&len[top.first-1][top.second]==-1)
37

{
38

len[top.first-1][top.second]=len[top.first][top.second]+1;
39

q.push(make_pair<int,int>(top.first-1,top.second));
40

}
41

if(top.first+1<=n&&used[top.first+1][top.second]&&len[top.first+1][top.second]==-1)
42

{
43

len[top.first+1][top.second]=len[top.first][top.second]+1;
44

q.push(make_pair<int,int>(top.first+1,top.second));
45

}
46

if(top.second-1>=0&&used[top.first][top.second-1]&&len[top.first][top.second-1]==-1)
47

{
48

len[top.first][top.second-1]=len[top.first][top.second]+1;
49

q.push(make_pair<int,int>(top.first,top.second-1));
50

}
51

if(top.second+1<=m&&used[top.first][top.second+1]&&len[top.first][top.second+1]==-1)
52

{
53

len[top.first][top.second+1]=len[top.first][top.second]+1;
54

q.push(make_pair<int,int>(top.first,top.second+1));
55

}
56

}
57

return len[er][ec]==-1?-1:len[er][ec]*10;
58

}
59

bool detect( int r, int c, int pos)
60

{
61

if(sta[pos].r==r||sta[pos].c==c) return true;
62

double dr=10*(r-sta[pos].r),dc=10*(c-sta[pos].c),dh=-sta[pos].h;
63

bool f1=r>sta[pos].r,f2=c>sta[pos].c;
64

//test row
65

for(int i=f1?sta[pos].r+1:sta[pos].r-1;f1?i<=r:i>=r;f1?i++:i--)
66

{
67

double k=10*(i-sta[pos].r)/dr,nc=sta[pos].c*10+k*dc,nh=sta[pos].h+k*dh;
68

int t1=f1?i-1:i,t2=f2?ceil(nc/10-eps)-1+eps:nc/10+eps;
69

if(gt(h[t1][t2],nh)) return false;
70

}
71

//test col
72

for(int i=f2?sta[pos].c+1:sta[pos].c-1;f2?i<=c:i>=c;f2?i++:i--)
73

{
74

double k=10*(i-sta[pos].c)/dc,nr=sta[pos].r*10+k*dr,nh=sta[pos].h+k*dh;
75

int t1=f1?ceil(nr/10-eps)-1+eps:nr/10+eps,t2=f2?i-1:i;
76

if(gt(h[t1][t2],nh)) return false;
77

}
78

return true;
79

}
80

int main()
81

{
82

int test;
83

scanf("%d",&test);
84

while(test--)
85

{
86

scanf("%d%d",&n,&m);
87

for(int i=0;i<n;i++)
88

for(int j=0;j<m;j++)
89

scanf("%d",&h[i][j]);
90

scanf("%d%d%d%d%d",&sr,&sc,&er,&ec,&num);
91

for(int i=0;i<num;i++)
92

scanf("%d%d%d",&sta[i].r,&sta[i].c,&sta[i].h);
93

//sort(sta,sta+num);
94

for(int i=0;i<=n;i++)
95

for(int j=0;j<=m;j++)
96

{
97

used[i][j]=false;
98

for(int k=0;k<num&&!used[i][j];k++)
99

if(detect(i,j,k)) used[i][j]=true;
100

}
101

/**//*printf("\n");
102

for(int i=0;i<=n;i++)
103

{
104

for(int j=0;j<=m;j++)
105

printf("%d ",used[i][j]);
106

printf("\n");
107

}*/
108

printf("%d\n",bfs());
109

}
110

return 0;
111

}

pku2902 Intercepting Missiles
这题我不知道说什么好，就是一个二分匹配模型，但是精度问题搞死我了，一直没能A，有A的请将代码贴在下面，万分感谢

pku2894-2903 Tehran 2005 比赛总结

pku2894-2903 Tehran 2005 比赛总结

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