hihor 学习日记：hiho一下 第五十三周（边的双连通分量）

http://hihocoder.com/contest/hiho53/problem/1

求出边的双联通分量

与求割点与割边类似，求边的双联通分量找的是桥，有m桥，那么就有m+1的分组。
而分组有点类似割边，如果一个分组中存在割边那么这个分组就不是分组，所以这张图的割边就是分组的边界，不包括在分组中，而low[u] == cur[u]正是割边所在的点，这是利用栈和DFS，把u后面入栈的点全部划分成同一分组，这样就写完了

AC代码：

#include <bits/stdc++.h>

using namespace std;
#define LL long long
const int Mod = 1e9 + 7;
const int maxn = 1e5 + 5;
const double eps = 0.00000001;
const int INF = 0x3f3f3f3f;

struct Edge{
    int v, next;
}edge[maxn << 1];

int tot, head[maxn];
int parent[maxn], vis[maxn], low[maxn], cur[maxn], trace[maxn];
int top, sta[maxn], cnt;

void init() {
    cnt = top = tot = 0;
    memset(head, -1, sizeof(head));
    memset(parent, -1, sizeof(parent));
    memset(vis, 0, sizeof(vis));
}

void addEdge(int u, int v) {
    edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot ++;

    edge[tot].v = u;
    edge[tot].next = head[v];
    head[v] = tot ++;
}

void DFS(int u) {
    static int counter = 0;
    int child = 0;
    vis[u] = 1;
    cur[u] = low[u] = ++counter;
    sta[++ top] = u;
    for (int i = head[u]; i + 1; i = edge[i].next) {
        int v = edge[i].v;
        if(!vis[v]) {
            child ++;
            parent[v] = u;
            DFS(v);
            low[u] = min(low[u], low[v]);
            if(cur[u] < low[v]) {
                cnt ++;
                //printf("bridge: %d %d\n", u, v);
            }
        }else if(v != parent[u])
            low[u] = min(low[u], cur[v]);
    }
    if(low[u] >= cur[u]) {
        int tt = top;
        int minn = u;
        while(sta[tt] != u) {
            minn = min(minn, sta[tt]);
            tt --;
        }
        while(sta[top] != u) {
            trace[sta[top]] = minn;
            top --;
        }
        if(sta[top] == u) {
            trace[u] = minn;
            top --;
        }
    }
}

int main()
{
    init();
    int N, M;
    cin >> N >> M;
    while(M --) {
        int u, v;
        cin >> u >> v;
        addEdge(u, v);
    }
    DFS(1);
    cout << cnt + 1<< endl;
    for (int i = 1; i <= N; i ++)
        cout << trace[i] << " ";
    cout << endl;
    return 0;
}