『算法-ACM 竞赛-图论』边双连通 V-DCC 缩点

图论–边双连通 V-DCC 缩点

// tarjan算法求无向图的割点、点双连通分量并缩点
#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], stack[SIZE], new_id[SIZE], c[SIZE];
int n, m, tot, num, root, top, cnt, tc;
bool cut[SIZE];
vector<int> dcc[SIZE];
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) {
    dfn[x] = low[x] = ++num;
    stack[++top] = x;
    if (x == root && head[x] == 0) { // 孤立点
        dcc[++cnt].push_back(x);
        return;
    }
    int flag = 0;
    for (int i = head[x]; i; i = Next[i]) {
        int y = ver[i];
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
            if (low[y] >= dfn[x]) {
                flag++;
                if (x != root || flag > 1) cut[x] = true;
                cnt++;
                int z;
                do {
                    z = stack[top--];
                    dcc[cnt].push_back(z);
                } while (z != y);
                dcc[cnt].push_back(x);
            }
        }
        else low[x] = min(low[x], dfn[y]);
    }
}

int main() {
    cin >> n >> m;
    tot = 1;
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        if (x == y) continue;
        add(x, y), add(y, x);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) root = i, tarjan(i);
    for (int i = 1; i <= n; i++)
        if (cut[i]) printf("%d ", i);
    puts("are cut-vertexes");
    for (int i = 1; i <= cnt; i++) {
        printf("v-DCC #%d:", i);
        for (int j = 0; j < dcc[i].size(); j++)
            printf(" %d", dcc[i][j]);
        puts("");
    }
    // 给每个割点一个新的编号(编号从cnt+1开始)
    num = cnt;
    for (int i = 1; i <= n; i++)
        if (cut[i]) new_id[i] = ++num;
    // 建新图，从每个v-DCC到它包含的所有割点连边
    tc = 1;
    for (int i = 1; i <= cnt; i++)
        for (int j = 0; j < dcc[i].size(); j++) {
            int x = dcc[i][j];
            if (cut[x]) {
                add_c(i, new_id[x]);
                add_c(new_id[x], i);
            }
            else c[x] = i; // 除割点外，其它点仅属于1个v-DCC
        }
    printf("缩点之后的森林，点数 %d，边数 %d\n", num, tc / 2);
    printf("编号 1~%d 的为原图的v-DCC，编号 >%d 的为原图割点\n", cnt, cnt);
    for (int i = 2; i < tc; i += 2)
        printf("%d %d\n", vc[i ^ 1], vc[i]);
}