hdu 1269 迷宮城堡（tarjan判連通）

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1269

本題是判斷連通分量個數是否爲1，裸題。

代碼：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;

const int maxn = 10005;
const int maxm = 100005;
int n, m;

int head[maxn];
struct E {
    int to;
    int next;
} edge[maxm];

int dfn[maxn];
int low[maxn];
bool instack[maxn];
int stap[maxn];

int index, cur, cnt, eid;

void addedge(int from, int to) {
    edge[eid].to = to;
    edge[eid].next = head[from];
    head[from] = eid ++;
}

void targan(int v) {
    if(cnt > 1) return ;
    dfn[v] = low[v] = ++index;
    instack[v] = true;
    stap[++cur] = v;
    for (int i = head[v]; ~i; i = edge[i].next) {
        int to = edge[i].to;
        if(!dfn[to]) {
            targan(to);
            low[v] = min(low[v], low[to]);
        } else if(instack[to]) {
            low[v] = min(low[v], dfn[to]);
        }
    }
    if(dfn[v] == low[v]) {
        cnt ++;
        if(cnt > 1) return ;
        int j;
        do {
            j = stap[cur--];
            instack[j] = false;
        } while (j != v);
    }
}
int main() {
    int x, y;
    while (~scanf("%d%d", &n, &m), m + n) {
        memset(head, -1, sizeof(head));
        memset(dfn, 0, sizeof(dfn));
        memset(instack, false, sizeof(instack));
        eid = 0;
        for (int i = 0; i < m; i ++) {
            scanf ("%d%d", &x, &y);
            addedge(x, y);
        }
        cur = index = cnt = 0;
        for (int i = 1; i <= n; i++) {
            if(!dfn[i]) {
                targan(i);
                if(cnt > 1) break;
            }
        }
        if(cnt == 1) {
            printf("Yes\n");
        } else {
            printf("No\n");
        }
    }
    return 0;
}