POJ2762 Going from u to v or from v to u? 强连通分量缩点+拓扑排序

题目链接：https://vjudge.net/contest/295959#problem/I 或者 http://poj.org/problem?id=2762

题意：输入多组样例,输入n个点和m条有向边,问该图中任意两点x, y之间是否满足x可以到y或者y可以到x。

一开始WA的原因是因为没注意到是或者, 如果是并且的话,就是一道简单的强连通分量的题,直接判断整个图是否为一个强连通分量

对于该题, 先用强连通分量进行缩点,简化图。图就变成了DAG,用拓扑排序判断图中点的入度, 图中入度为0的点只能存在一个, 若存在2个及以上的话,那么这2个点或多个点是无法互相到达。

在拓扑排序中用列队, 入度为0的点入队,每次都对列队中的数判断,若大于等于2则直接return 不满足。

#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;

int n, m, cnt, cnt1;
int head[1010], head1[1010];
int dfn[1010], low[1010], vis[1010], deep, k_color, color[1010];
stack<int>S;
int in[1010], flag;

struct Edge
{
	int to, next;
}edge[6010], edge1[6010];

void add(int a, int b)
{
	edge[++ cnt].to = b;
	edge[cnt].next = head[a];
	head[a] = cnt;
}

void add1(int a, int b)
{
	edge[++ cnt1].to = b;
	edge[cnt1].next = head1[a];
	head1[a] = cnt1;
}

void tarjan(int now)
{
	dfn[now] = low[now] = ++ deep;
	vis[now] = 1;
	S.push(now);
	for(int i = head[now]; i != -1; i = edge[i].next)
	{
		int to = edge[i].to;
		if(!dfn[to])
		{
			tarjan(to);
			low[now] = min(low[now], low[to]);
		}
		else if(vis[to])
		{
			low[now] = min(low[now], dfn[to]);
		}
	}
	if(dfn[now] == low[now])
	{
		k_color ++;
		while(1)
		{
			int temp = S.top();
			S.pop();
			color[temp] = k_color;
			vis[temp] = 0;
			if(temp == now)
				break;
		}
	}
}

void topo_sort()
{
	queue<int>Q;
	while(!Q.empty())
		Q.pop();
	for(int i = 1; i <= k_color; i ++)
		if(!in[i])
			Q.push(i);
	if(Q.size() > 1)
	{
		flag = 0;
		return ;
	}
	while(!Q.empty())
	{
		int temp = Q.front();
		Q.pop();
		for(int i = head1[temp]; i != -1; i = edge[i].next)
		{
			int to = edge[i].to;
			in[to] --;
			if(!in[to])
				Q.push(to);
			if(Q.size() > 1)
			{
				flag = 0;
				return ;
			}
		}
	}
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T --)
	{
		flag = 1;
		cnt = deep = k_color = cnt1 = 0;
		memset(head, -1, sizeof(head));
		memset(in, 0, sizeof(in));
		memset(head1, -1, sizeof(head1));
		memset(dfn, 0, sizeof(dfn));
		memset(low, 0, sizeof(low));
		memset(vis, 0, sizeof(vis));
		scanf("%d%d", &n, &m);
		for(int i = 1; i <= m; i ++)
		{
			int a, b;
			scanf("%d%d", &a, &b);
			add(a, b);
		}
		for(int i = 1; i <= n; i ++)
			if(!dfn[i])
				tarjan(i);
		for(int i = 1; i <= n; i ++)//缩点
		{
			for(int j = head[i]; j != -1; j = edge[j].next)
			{
				int to = edge[j].to;
				int x = color[i], y = color[to];
				if(x != y)
				{
					add1(x, y);
					in[y] ++;
				}
			}
		}
		topo_sort();
		if(flag)
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}

原文地址：https://www.cnblogs.com/yuanweidao/p/10757348.html