POJ 2186 Popular Cows (強聯通分量)

標籤：

連結 ：http://poj.org/problem?id=2186

一個聯通分量裡的所有的牛滿足任何一個被其他牛認為是紅人。強聯通縮點之後 只需要找到一個且只有一個聯通分量且它的出度為0 答案就是這個強聯通分量點的個數。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <stack>#include <cmath>#include <map>#define lson o<<1,l,m#define rson o<<1|1,m+1,r#define mem(a) memset(a,0,sizeof(a))typedef long long ll;const int N = 10005;const int M = 50005;const ll mod = 1000000007;using namespace std;int n, m, dfs_clock, scc_cnt;int he[N], pre[N], low[N], scc[N];stack <int> S;struct C {    int ne, to;} e[M];void add(int id, int x, int y) {    e[id].to = y;    e[id].ne = he[x];    he[x] = id;}void dfs(int u) {    pre[u] = low[u] = ++ dfs_clock;    S.push(u);    for(int i = he[u]; i != -1; i = e[i].ne) {        int v = e[i].to;        if(pre[v] == 0) {            dfs(v);            low[u] = min(low[u], low[v]);        } else if(scc[v] == 0) {            low[u] = min(low[u], pre[v]);        }    }    if(low[u] == pre[u]) {        scc_cnt ++;        while(1) {            int x = S.top(); S.pop();            scc[x] = scc_cnt;            if(x == u) break;        }    }}void find_scc() {    mem(scc);    mem(pre);    dfs_clock = scc_cnt = 0;    for(int i = 1; i <= n; i++) {        if(pre[i] == 0) dfs(i);    }}int out[N], v[N];int main() {    while(cin >> n >> m) {        memset(he, -1, sizeof(he));        for(int i = 1; i <= m; i++) {            int x, y;            scanf("%d%d", &x, &y);            add(i, x, y);        }        find_scc();        mem(v);        for(int i = 1; i <= n; i++) {            v[ scc[i] ]++;        }        //printf("%d\n", scc_cnt);        int ans;        for(int u = 1; u <= n; u++) {            for(int i = he[u]; i != -1; i = e[i].ne) {                int v = e[i].to;                if(scc[u] != scc[v]) {                    out[scc[u]] = 1;                }            }        }        int cnt = 0;        for(int i = 1; i <= scc_cnt; i++) {            if(out[i] == 0) {                cnt++;                ans = v[i];            }        }        if(cnt == 1) {            printf("%d\n", ans);        } else puts("0");    }    return 0;}

 

POJ 2186 Popular Cows (強聯通分量)