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  • tarjan算法求无向图的桥、边双连通分量并缩点

    // tarjan算法求无向图的桥、边双连通分量并缩点
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    const int SIZE = 100010;
    int head[SIZE], ver[SIZE * 2], Next[SIZE * 2];
    int dfn[SIZE], low[SIZE], c[SIZE];
    int n, m, tot, num, dcc, tc;
    bool bridge[SIZE * 2];
    int hc[SIZE], vc[SIZE * 2], nc[SIZE * 2];
    
    void add(int x, int y) {
        ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
    }
    
    void add_c(int x, int y) {
        vc[++tc] = y, nc[tc] = hc[x], hc[x] = tc;
    }
    
    void tarjan(int x, int in_edge) {
        dfn[x] = low[x] = ++num;
        for (int i = head[x]; i; i = Next[i]) {
            int y = ver[i];
            //当前点未走过
            if (!dfn[y]) {
                tarjan(y, i);
                low[x] = min(low[x], low[y]);
                if (low[y] > dfn[x])
                    //i 与 i^1是桥
                    bridge[i] = bridge[i ^ 1] = true;
            }
            //反向边更新
            else if (i != (in_edge ^ 1))
                low[x] = min(low[x], dfn[y]);
        }
    }
    
    void dfs(int x) {
        c[x] = dcc;
        for (int i = head[x]; i; i = Next[i]) {
            int y = ver[i];
            if (c[y] || bridge[i]) continue;
            dfs(y);
        }
    }
    
    int main() {
        cin >> n >> m;
        tot = 1;
        for (int i = 1; i <= m; i++) {
            int x, y;
            scanf("%d%d", &x, &y);
            add(x, y), add(y, x);
        }
        for (int i = 1; i <= n; i++)
            if (!dfn[i]) tarjan(i, 0);
        for (int i = 2; i < tot; i += 2)
            if (bridge[i])//当前桥,输出连通桥的两点
                printf("%d %d
    ", ver[i ^ 1], ver[i]);
            //求边双连通分量(不存在桥)
        for (int i = 1; i <= n; i++)
            if (!c[i]) {
                ++dcc;
                dfs(i);
            }
        printf("There are %d e-DCCs.
    ", dcc);
        for (int i = 1; i <= n; i++)
            printf("%d belongs to DCC %d.
    ", i, c[i]);
    
        //缩点
        tc = 1;
        for (int i = 2; i <= tot; i++) {
            int x = ver[i ^ 1], y = ver[i];
            if (c[x] == c[y]) continue;
            add_c(c[x], c[y]);
        }
        printf("缩点之后的森林,点数 %d,边数 %d
    ", dcc, tc / 2);
        for (int i = 2; i < tc; i += 2)
            printf("%d %d
    ", vc[i ^ 1], vc[i]);
    }
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  • 原文地址:https://www.cnblogs.com/DWVictor/p/11347505.html
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