hdu 5409 CRB and Graph（边双联通分量）

题意：

给一个图一些边，保证图连通

问对于每条边，如果去除该边后使得图中一些点不连通。设这些点（u，v），要求使u尽量小，v尽量大，输出这样的（u，v）。否则输出0 0。

 #include <bits/stdc++.h>
 using namespace std;
 const int MAXN = 1e5 + ;
 typedef pair <int, int>pii;
 vector<pii>G[MAXN];
 bool isBridge[MAXN];
 int clk, pre[MAXN], low[MAXN];
 int IDX, maxv[MAXN], newIdx[MAXN], newMax[MAXN];
 int U[MAXN], V[MAXN];
 int ans[MAXN];
 bool vis[MAXN];
 int n, m;

 void init () {
     memset(isBridge, false, sizeof (isBridge)); //记录桥
     memset(pre, , sizeof (pre));   //记录的第一次访问的时间戳
     memset(low, , sizeof (low));   //本身及其子节点能回到的最早的祖先的pre值
     clk = ;    //时间戳
     for (int i = ; i < MAXN; i++) {
         G[i].clear();
     }
 }

 void DFS (int u, int pa) {
     int lowu = pre[u] = ++clk;
     for (int i = ; i < G[u].size(); i++) {
         pii e = G[u][i];
         int v = e.first;
         int idx = e.second;
         if (!pre[v]) {
             DFS(v, u);
             lowu = min(lowu, low[v]);
             if (low[v] > pre[u]) {
                 isBridge[idx] = true;
             }
         } else if (pre[v] < pre[u] && v != pa) {
             //是反向边更新lowu
             lowu = min(lowu, pre[v]);
         }
     }
     low[u] = lowu;  //更改low[u]
 }

 void DFS2(int u, int pa) {
     vis[u] = true;
     maxv[u] = u;
     newIdx[u] = IDX;
     for (int i = ; i < G[u].size(); i++) {
         pii e = G[u][i];
         int v = e.first;
         int idx = e.second;
         if (!isBridge[idx] && v != pa && !vis[v]) {
             DFS2(v, u);
             maxv[u] = max(maxv[u], maxv[v]);
         }
     }
 }

 void BCC_Bridge() {
     DFS(, -); //记录桥
     memset(vis, false, sizeof (vis));
     IDX = ;
     for (int i = ; i <= n; i++) {
         if (!vis[i]) {
             IDX++;
             DFS2(i, -);    //缩点
         }
     }
     //重新记录缩后的点
     for (int i = ; i <= n; i++) {
         G[i].clear();
     }
     for (int i = ; i < m; i++) {
         if (isBridge[i]) {
             int u = newIdx[U[i]], v = newIdx[V[i]];
             G[u].push_back(make_pair(v, i));
             G[v].push_back(make_pair(u, i));
         }
     }
 }

 void solve (int u, int pa) {
     pre[u] = ++clk;
     ans[u] = newMax[u];
     for (int i = ; i < G[u].size(); i++) {
         int v = G[u][i].first;
         if (v != pa) {
             solve(v, u);
             ans[u] = max(ans[u], ans[v]);
         }
     }
 }

 int main() {
     int T;
     scanf ("%d", &T);
     while (T--) {
         init(); //进行初始化
         scanf ("%d%d", &n, &m);
         for (int i = ; i < m; i++) {
             int u, v;
             scanf ("%d%d", &u, &v);
             U[i] = u, V[i] = v;
             G[u].push_back(make_pair(v, i));
             G[v].push_back(make_pair(u, i));
         }
         BCC_Bridge();
         for (int i = ; i <= n; i++) {
             newMax[newIdx[i]] = maxv[i];
         }
         int u;
         for (u = ; u <= n; u++) {
             if (newMax[u] == n) {
                 break;
             }
         }
         memset(pre, , sizeof pre);
         clk = ;    //重新定义时间戳
         solve (u, );
         for (int i = ; i < m; i++) {
             int u = newIdx[U[i]],  v = newIdx[V[i]];
             if (u == v) {
                 printf("0 0\n");
             } else {
                 if (pre[u] < pre[v]) {
                     swap(u, v);
                 }
                 printf("%d %d\n", ans[u], ans[u]+);
             }
         }
     }
     return ;
 }