poj2186 Popular Cows
2013-09-21 12:08
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Popular Cows
Description
Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
Input
* Line 1: Two space-separated integers, N and M
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
Output
* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
Sample Input
Sample Output
Hint
Cow 3 is the only cow of high popularity.
Source
USACO 2003 Fall
先求强连通分量,缩点之后,如果出度为0的只有一个,那个结果自然是这个缩点的点数,如果为0的大于1,则一定不连通,所以直接输出0就可以了!
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 20718 | Accepted: 8438 |
Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
Input
* Line 1: Two space-separated integers, N and M
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
Output
* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
Sample Input
3 3 1 2 2 1 2 3
Sample Output
1
Hint
Cow 3 is the only cow of high popularity.
Source
USACO 2003 Fall
先求强连通分量,缩点之后,如果出度为0的只有一个,那个结果自然是这个缩点的点数,如果为0的大于1,则一定不连通,所以直接输出0就可以了!
#include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #define N 10510 #define E 100000 #define M 1000000 using namespace std; int head ,sta[M],num ,re,ans,next[E],id ,in ,vec[E],vis ,dfn ,low ,clock_m,edge_m; int addedge(int s,int e){ vec[edge_m]=e;next[edge_m]=head[s];head[s]=edge_m++; } int init(){ memset(vis,0,sizeof(vis)); memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(num,0,sizeof(num)); memset(head,-1,sizeof(head)); memset(id,0,sizeof(id)); memset(in,0,sizeof(in)); edge_m=0;clock_m=0;ans=0;re=0; } int tarjan(int x){ dfn[x]=low[x]=++clock_m; sta[++ans]=x; vis[x]=1; for(int i=head[x];i!=-1;i=next[i]){ int goal=vec[i]; if(!dfn[goal]){ tarjan(goal); low[x]=min(low[x],low[goal]); } else if(/*vis[goal]*/!id[goal]) low[x]=min(low[x],dfn[goal]); } if(low[x]==dfn[x]){ re++;int v; do{ v=sta[ans--];vis[v]=0;num[re]++;id[v]=re; }while(v!=x); } return 1; } int main() { int tcase,n,m,s,e; while(scanf("%d%d",&n,&m)!=EOF) { //system("PAUSE"); init(); for(int i=0;i<m;i++){ scanf("%d%d",&s,&e); addedge(s,e); } for(int i=1;i<=n;i++) if(!dfn[i]) { tarjan(i); } for(int i=1;i<=n;i++){ for(int j=head[i];j!=-1;j=next[j]){ if(id[vec[j]]!=id[i]){ in[id[i]]++; } } } ans=0;int k=1; for(int i=1;i<=re;i++){ if(in[i]==0){ sta[ans++]=i;k=i; } } if(ans>1)printf("0\n"); else printf("%d\n",num[k]); } return 0; }
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