首页 > itarticle > Tarjan无向图求桥、e-DCC求割点和v-DCC强联通分量及缩点 求割点和v-DCC 强联通分量及缩点

Tarjan无向图求桥、e-DCC求割点和v-DCC强联通分量及缩点 求割点和v-DCC 强联通分量及缩点

admin 11月 11, 2020 0 

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#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 (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];
		if (!dfn[y]) {
			tarjan(y, i);
			low[x] = min(low[x], low[y]);
			if (low[y] > dfn[x])
				bridge[i] = bridge[i ^ 1] = true;
		}
		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])
			printf("%d %dn", ver[i ^ 1], ver[i]);

	for (int i = 1; i <= n; i++)
		if (!c[i]) {
			++dcc;
			dfs(i);
		}
	printf("There are %d e-DCCs.n", dcc);
	for (int i = 1; i <= n; i++)
		printf("%d belongs to DCC %d.n", i, c[i]);

	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 %dn", dcc, tc / 2);
	for (int i = 2; i < tc; i += 2)
		printf("%d %dn", vc[i ^ 1], vc[i]);
}

求割点和v-DCC

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#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 (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) { 
		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 %dn", 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 %dn", vc[i ^ 1], vc[i]);
}

强联通分量及缩点

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int N = 100010, M = 1000010;
int ver[M], Next[M], head[N], dfn[N], low[N];
int stack[N], ins[N], c[N];
int vc[M], nc[M], hc[N], tc;
vector<int> scc[N];
int n, m, tot, num, top, cnt;

void (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, ins[x] = 1;
	for (int i = head[x]; i; i = Next[i])
		if (!dfn[ver[i]]) {
			tarjan(ver[i]);
			low[x] = min(low[x], low[ver[i]]);
		}
		else if (ins[ver[i]])
			low[x] = min(low[x], dfn[ver[i]]);
	if (dfn[x] == low[x]) {
		cnt++; int y;
		do {
			y = stack[top--], ins[y] = 0;
			c[y] = cnt, scc[cnt].push_back(y);
		} while (x != y);
	}
}

int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		add(x, y);
	}
	for (int i = 1; i <= n; i++)
		if (!dfn[i]) tarjan(i);
	for (int x = 1; x <= n; x++)
		for (int i = head[x]; i; i = Next[i]) {
			int y = ver[i];
			if (c[x] == c[y]) continue;
			add_c(c[x], c[y]);
		}
}
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