首页 > 其他 > 详细

T103492 【模板】点双连通分量

时间:2019-10-26 11:50:36      阅读:101      评论:0      收藏:0      [点我收藏+]

技术分享图片

技术分享图片

题目地址 


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int SIZE = 1e5;
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;
vector<int> dcc[SIZE];
void add(int x, int y) {
	ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}
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;
	}
	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]) {
				//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 <= cnt; i++) {
		for (int j = 0; j < dcc[i].size(); j++)
			printf("%d ", dcc[i][j]);
		puts("");
	}
	return 0;
}

  

 

T103492 【模板】点双连通分量

原文:https://www.cnblogs.com/zbsy-wwx/p/11742089.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!