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 + 1;
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, 0, sizeof (pre));   //记录的第一次访问的时间戳
    memset(low, 0, sizeof (low));   //本身及其子节点能回到的最早的祖先的pre值
    clk = 0;    //时间戳
    for (int i = 0; i < MAXN; i++) {
        G[i].clear();
    }
}

void DFS (int u, int pa) {
    int lowu = pre[u] = ++clk;
    for (int i = 0; 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 = 0; 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(1, -1); //记录桥
    memset(vis, false, sizeof (vis));
    IDX = 0;
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            IDX++;
            DFS2(i, -1);    //缩点
        }
    }
    //重新记录缩后的点
    for (int i = 1; i <= n; i++) {
        G[i].clear();
    }
    for (int i = 0; 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 = 0; 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 = 0; 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 = 1; i <= n; i++) {
            newMax[newIdx[i]] = maxv[i];
        }
        int u;
        for (u = 1; u <= n; u++) {
            if (newMax[u] == n) {
                break;
            }
        }
        memset(pre, 0, sizeof pre);
        clk = 0;    //重新定义时间戳
        solve (u, 0);
        for (int i = 0; i < m; i++) {
            int u = newIdx[U[i]],  v = newIdx[V[i]];
            if (u == v) {
                printf("0 0
");
            } else {
                if (pre[u] < pre[v]) {
                    swap(u, v);
                }
                printf("%d %d
", ans[u], ans[u]+1);
            }
        }
    }
    return 0;
}