BNU Training 2015 07 27 题解
【比赛链接】：click here~~
uva 12435 C. Consistent Verdicts
【题目大意】：给你二维平面一些人的坐标，每个人手上都有一把枪，求全部人同时开枪后所有人听到枪声的次数的可能数目。【解题思路】：O(n^2)暴力枚举+unique 函数去重相邻元素。居然只跑了3ms，~~
代码：
<span style="font-size:14px;">// C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
using namespace std;
#define rep(i,j,k) for(int i=(int)j;i<(int)k;++i)
#define per(i,j,k) for(int i=(int)j;i>(int)k;--i)
#define lowbit(a) a&-a
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef unsigned long long LLU;
typedef double db;
const int N=1e6+10;
const int inf=0x3f3f3f3f;
char str[N];
bool vis[N];
int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
int movv[5][2]= {{1,0},{0,1},{0,0},{-1,0},{0,-1}};
inline LL read()
{
int c=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
c=c*10+ch-'0';
ch=getchar();
}
return c*f;
}
char mon1[N],mon2[N];
LL day1,year1;
LL day2,year2;
LL row,line,x,t,y,i,res;
struct node
{
LL codrx;
LL codry;
};
node num[N];
LL mum[N];
int main()
{
LL tot=1;
t=read();
while(t--){
x=read();
int len=0;
for(i=0; i<x; ++i){
num[i].codrx=read();
num[i].codry=read();
}
for(int i=0; i<x; ++i){
for(int j=i+1; j<x; ++j)
mum[len++]=(num[i].codrx-num[j].codrx)*(num[i].codrx-num[j].codrx)+(num[i].codry-num[j].codry)*(num[i].codry-num[j].codry);
}
sort(mum,mum+len);
LL res = unique(mum,mum+len)-mum;
printf("Case %lld: %lld\n",tot++,res+1);
}
return 0;
}</span>
uva 12439 G. February 29
【题目大意】两个日期之间求leap data的个数：ps:常规方法判断会TE，因此换个思路，判断闰年可以用除法
代码：
<span style="font-size:14px;">// C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
using namespace std;
#define rep(i,j,k) for(int i=(int)j;i<(int)k;++i)
#define per(i,j,k) for(int i=(int)j;i>(int)k;--i)
#define lowbit(a) a&-a
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef unsigned long long LLU;
typedef double db;
const int N=1e5;
const int inf=0x3f3f3f3f;
char str[N];
bool vis[N];
int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
int movv[5][2]= {{1,0},{0,1},{0,0},{-1,0},{0,-1}};
bool leap_year(int y)
{
if(y%4==0&&y%100!=0||y%400==0) return 1;
return 0;
}
inline LL read()
{
int c=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
c=c*10+ch-'0';
ch=getchar();
}
return c*f;
}
char mon1[N],mon2[N];
LL day1,year1;
LL day2,year2;
LL row,line,x,t,y,i,res;
int main()
{
scanf("%lld",&t);
for(i=1; i<=t; ++i)
{
int res=0;
scanf("%s %lld,%lld",mon1,&day1,&year1);
scanf("%s %lld,%lld",mon2,&day2,&year2);
if((mon1[0]=='J'&&mon1[1]=='a')||mon1[0]=='F') year1=year1;
else year1++;
if((mon2[0]=='J'&&mon2[1]=='a')||mon2[0]=='F'&&day2<=28) year2--;
res=(year2/4)-((year1-1)/4);
res=res-(year2/100)+((year1-1)/100);
res=res+(year2/400)-((year1-1)/400);
printf("Case %lld: %lld\n",i,res);
}
return 0;
}</span>
uva 12442 J. Forwarding Emails
【题目大意】：酋长发一封信给部落的人，给出部落的人的关系，没人只能收一次且发一次，求在信能传递的最长的的人数链的情况下第一次收到信的人的编号
【解题思路】：
网上看到一个比较直观的思路过程，复制过来，便于理解：
Solution Description :
DFS problem
And count_visited_node=2 (1->2)
And count_visited_node=3 (3->2->1)
And count_visited_node=4 (4->3->2->1)
代码：
DFS problem
Read this line carefully in problem description - "they each pick one other person they know to email those things to every time - exactly one, no less, no more (and never themselves)"
i.e Each person send email only one. For each person has only one adjacency person. The input can not be (1 to 3, 2 to 3, 1 to 2) , because for person 1 here 2 adjacency person 3 and 2.
So, you can represent this graph using one dimensional array adj[N].
Do not need to stack, you can use recursion.
Example:
4
1 2
2 1
4 3
3 2
The adjacency list, adj[1]=2, adj[2]=1, adj[3]=2, and adj[4]=3.
Use a boolean array visit[N]
visit[1] |
visit[2] |
visit[3] |
visit[4] |
False |
False |
False |
False |
At first start from 1
If visit[1]==false then run DFS in this time count the visited node and update the visit[] array (Set visit[i]=True here i is a visited node). Remember dot not use the array visit[] for cycle finding you can use another boolean array visit2[] for that purpose. After run DFS the visit[] array is
visit[1] |
visit[2] |
visit[3] |
visit[4] |
True |
True |
False |
False |
Now 2
visit[2]==true so, do not need to do anything.
Now 3
visit[3]==false so, run the DFS. After that the visit[] array is
visit[1] |
visit[2] |
visit[3] |
visit[4] |
True |
True |
True |
False |
Now 4visit[4]==false so, run the DFS. After that the visit[] array is
visit[1] |
visit[2] |
visit[3] |
visit[4] |
True |
True |
True |
True |
For 4, the count_visited_node is maximum. Ans is 4.
<span style="font-size:14px;">#include <stdio.h>
#define MAX 50005
int T,N;
int vis[MAX], f[MAX], c[MAX];
int ans, flag;
typedef long long LL;
inline LL read()
{
int c=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
return c*f;
}
int dfs(int u)
{
int v = f[u];/// 2=f[1];1=f[2];2=f[3];
int r = 0; ///
vis[u] = 1; /// vis[1]=1;vis[2]=1;vis[3]=1;
if(!vis[v]) r = dfs(v) + 1;///vis[2]=1,r=0+1=1;
vis[u] = 0;
c[u] = r; ///c[1]=1,c[2]=0,c[3]=2;
return r;
}
int main()
{
int u,v;
T=read();
for(int t=1; t<=T; t++){
N=read();
for(int i=1; i<=N; i++){
u=read(),v=read();
f[u] = v;
vis[u] = 0;
c[u] = -1;
}
ans = -1;
for(int i=1; i<=N; i++){
if(c[i]==-1) dfs(i);
if(c[i]>m)
{
m=c[i];
flag=i;
}
/* printf("vertex %d children %d\n",i,c[i]);*/
}
printf("Case %d: %d\n",t,flag);
}
return 0;
}</span>
7
3
1 2
2 3
3 1
4
1 2
2 1
4 3
3 2
5
1 2
2 1
5 3
3 4
4 5
2
1 2
2 1
3
1 2
2 3
3 1
4
4 2
2 1
4 3
3 2
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 1
Case 1: 1
Case 2: 4
Case 3: 3
Case 4: 1
Case 5: 1
Case 6: 4
Case 7: 1