洛谷 [P2341] 受欢迎的牛

强连通分量

一个结论: 在有向图中, 一个联通块能被所有点遍历当且仅当图中只有一个连通块出度为零

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
using namespace std;
const int MAXN = 100005;
int init() {
    int rv = 0, fh = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fh = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        rv = (rv<<1) + (rv<<3) + c - '0';
        c = getchar();
    }
    return fh * rv;
}
int n, m, nume, head[MAXN], dfn[MAXN], low[MAXN], ind, cnt, fa[MAXN], num[MAXN], out[MAXN], ans;
bool f[MAXN];
struct edge{
    int to, nxt;
}e[MAXN];
void adde(int from, int to) {
    e[++nume].to = to;
    e[nume].nxt = head[from];
    head[from] = nume;
}
stack <int> sta;
void tarjan(int u) {
    low[u] = dfn[u] = ++ind;
    f[u] = 1;
    sta.push(u);
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }else if(f[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]) {
        int v;
        cnt++;
        do {
            v = sta.top();
            fa[v] = cnt;
            f[v] = 0;
            num[cnt]++;
            sta.pop();
        } while(v != u);
    }
}
int main() {
    n = init(); m = init();
    for(int i = 1; i <= m; i++) {
        int u = init(), v = init();
        adde(u, v);
    }
    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; j = e[j].nxt) {
            int v = e[j].to;
            if(fa[i] != fa[v]) out[fa[i]]++;
        }
    }
    for(int i = 1; i <= cnt; i++) {
        if(!out[i]) {
            if(ans) {ans = 0;break;}
            ans = num[i];
        }
    }
    cout << ans << endl;
    return 0;
}
posted @ 2018-05-24 16:23  Mr_Wolfram  阅读(...)  评论(...编辑  收藏