题意：找无向图的桥。很长时间没敲过图论代码了，这个算法居然忘了。
解法：pre[x]记录顶点x的访问时间，anc[x]记录通过顶点x的dfs子树所能访问到的pre最小的祖先结点的访问时间。如果 anc[x]==pre[x] 说明到达顶点x的树边是桥。
#include <stdio.h>
#include <memory.h>
#include <vector>
#include <algorithm>
using namespace std;const int maxn = 10005;
const int maxm = 100005 * 2;struct Graph {int hed[maxn], pnt[maxm], nxt[maxm], id[maxm], cnt;void init(int n) {memset(hed + 1, -1, 4 * n);cnt = 0;}void addedge(int x, int y, int i) {pnt[cnt] = y; id[cnt] = i; nxt[cnt] = hed[x]; hed[x] = cnt++;pnt[cnt] = x; id[cnt] = i; nxt[cnt] = hed[y]; hed[y] = cnt++;}
} G;vector<int> V;
int anc[maxn], pre[maxn], vis[maxn];void dfs(int x, int f, int d) {vis[x] = 1;anc[x] = pre[x] = d;for (int p = G.hed[x]; p != -1; p = G.nxt[p]) {int y = G.pnt[p];int e = G.id[p];if (e != f && vis[y]) { // back edgeif (anc[x] > pre[y])anc[x] = pre[y];} else if (!vis[y]) {dfs(y, e, d + 1); // tree edgeif (anc[y] < anc[x])anc[x] = anc[y];if (anc[y] > pre[x])V.push_back(e);}}
}int main() {int T, N, M, i, x, y;for (scanf("%d", &T); T--; ) {scanf("%d %d", &N, &M);G.init(N);for (i = 1; i <= M; i++) {scanf("%d %d", &x, &y);G.addedge(x, y, i);}memset(vis + 1, 0, 4 * N);V.clear();dfs(1, -1, 0);sort(V.begin(), V.end());printf("%d\n", V.size());for (i = 0; i < V.size(); i++) {if (i) printf(" ");printf("%d", V[i]);}if (V.size()) printf("\n");if (T) printf("\n");}return 0;
}