|
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 3 3 1 2 2 1 2 3 Sample Output 1 Hint
Cow 3 is the only cow of high popularity.
Source |
分析：刚开始理解错题意了，以为是找出最受欢迎的牛，结果发现不是。就是找到每头牛都觉得该头牛牛x的个数。也就是在一个DAG中找出出度为零的点，如果有两个出度为零的点，则不可能有这样的一头牛。
方法：一、求强连通分支，将图转为DAG
二、在DAG中找到出度为零的连通分支或点。
代码：
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 10010;
vector<int> G[maxn],Gt[maxn];
int ord[maxn];
int vis[maxn];
int out[maxn];
int ss[maxn];
int ans[maxn];
int cnt;
int t;
int n,m;
void dfs_1(int u)
{
vis[u] = 1;
for(int i = 0; i < G[u].size(); i++)
{
int v = G[u][i];
if(!vis[v]) dfs_1(v);
}
ord[t++] = u;
}
void dfs_2(int u)
{
vis[u] = 1;
ss[u] = cnt;
for(int i = 0; i < Gt[u].size(); i++)
{
int v = Gt[u][i];
if(!vis[v])
{
ans[cnt]++;
dfs_2(v);
}
}
}
void kosaraju()
{
int flag = -1;
t = 1;
cnt = 1;
for(int i = 1; i <= n; i++) ans[i] = 1;
memset(out,0,sizeof(out));
memset(vis,0,sizeof(vis));
for(int i = 1; i <= n; i++) if(!vis[i]) dfs_1(i);
memset(vis,0,sizeof(vis));
for(int i = t - 1; i >= 1; i--)
{
int v = ord[i];
if(!vis[v])
{
dfs_2(v);
cnt++;
}
}
//构造DAG
for(int i = 1; i <= n; i++)
{
for(int j = 0; j < G[i].size(); j++)
{
if(ss[i] == ss[G[i][j]]) continue;
out[ss[i]]++;
}
}
//查找出度为零的点或者连通分支
for(int i = 1; i < cnt; i++)
{
if(!out[i] && flag == -1) flag = ans[i];
else if(!out[i] && flag != -1)
{
flag = 0;
break;
}
}
printf("%d\n",flag);
}
int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
for(int i = 1; i <= m; i++)
{
int u,v;
scanf("%d%d",&u,&v);
G[u].push_back(v);
Gt[v].push_back(u);
}
kosaraju();
}
return 0;
}