【luogu P2863 [USACO06JAN]牛的舞会The Cow Prom】 题解

标签：==   size   强连通   int   include   ems   set   lse   algo   

题目链接：https://www.luogu.org/problemnew/show/P2863
求强连通分量大小>自己单个点的

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
struct edge{
    int next, to, from, len;
}e[maxn<<2];
int head[maxn], cnt;
int n, m, color[maxn], tong[maxn], num, tim, ans;
int dfn[maxn], low[maxn];
bool vis[maxn];
stack<int> s;
void add(int u, int v)
{
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt;
}
void tarjan(int x)
{
    dfn[x] = low[x] = ++tim;
    s.push(x); vis[x] = 1;
    for(int i = head[x]; i != -1; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(vis[v])
        {
            low[x] = min(low[x], low[v]);
        }
    }
    if(dfn[x] == low[x])
    {
        color[x] = ++num;
        vis[x] = 0;
        while(s.top() != x)
        {
            color[s.top()] = num;
            vis[s.top()] = 0;
            s.pop(); 
        }
        s.pop();
    }
}
int main()
{
    memset(head, -1, sizeof(head));
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i++)
        tong[color[i]]++;
    for(int i = 1; i <= n; i++)
        if(tong[i] > 1) ans++;
    printf("%d",ans);
    return 0;
}

【luogu P2863 [USACO06JAN]牛的舞会The Cow Prom】 题解

标签：==   size   强连通   int   include   ems   set   lse   algo   

原文地址：https://www.cnblogs.com/MisakaAzusa/p/9368270.html