Road

最新推荐文章于 2024-10-18 08:00:00 发布

Seattle1

最新推荐文章于 2024-10-18 08:00:00 发布

阅读量411

点赞数

分类专栏：图论

图论专栏收录该内容

6 篇文章

订阅专栏

Road

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 23, Accepted users: 20

Problem 12934 : No special judgement

Problem description

In one country, there are N cities, numbered 1 to N. A civil engineer has to build public roads that connect all the cities together, i.e., it must be possible to travel from all cities to any other cities, maybe going through multiple cities. His team has surveyed several routes (candidate road between any two cities). Each route is a bi- directional connection between two cities. He can build a bi-directional road on the surveyed route for a specific cost (The shorter the route is the cheaper the road).

This engineer has never planned a road system in advance. He would just pick one of the routes based on his preference, and build a road until all the cities are connected.

Right now this engineer is going to build a road from the city p to the city q. With pressure from the government to reduce the cost, he asks you to write a program to decide, if he should build this road or not. Your program should say yes if the building of this road guaranteed that it can be part of the shortest road system that connects all cities together. Otherwise, your program should say no.

Input

First line of input is a number of test cases T ≤ 10.

Each test case start with a line contains 4 integers N, M, p, and q.

N (2 ≤ N ≤ 10 000) is the number of cities in this road system.

M (1 ≤ M ≤ 20 000) is the number of surveyed routes.

p and q (1 ≤ p ≤ N and 1 ≤ q ≤ N) indicate the route between two cities,

that the engineer asks if he can build a road or not. Then, the each of the following M lines contains 3 numbers u, v, and w (1 ≤ u ≤ N, 1 ≤ v ≤ N, 1 ≤ w ≤ 400 000) indicates that there are the bidirectional route of length w between u and v. The length of each road in the current system is unique. And, there is only one possible road between two cities. The input guarantees that at least one road system has a route between any two cities. All numbers are integer.

Output

For each test case, display YES if the engineer can build a road p-q as part of the shortest road system. Otherwise, print NO.

Sample Input

Sample Output

YES
NO
YES

#include<iostream>//不想说了，我再也不用模版了，自己做自己想 
#include<cstring>//加入变后合并这两个点 
#include<cstdlib>//8.24.ROAD 
#include<algorithm>
#include<cstdio>
#define MAXM 40010
#define MAXN 20010
using namespace std;
bool flag;
int n,m,strx,stry,strnum,k,ans;
struct EDGE
{
	int u,v,w;
}edge[MAXM];

bool cmp(EDGE a,EDGE b)
{
	return a.w<b.w;
}

void read(int *x)
{
    char ch;
    (*x) = 0;
    while((ch = getchar()) > 47 && ch < 58)
    {
        (*x) *= 10;
        (*x) += ch - '0';
    }
}



void union_point(int a,int b)//??
{
	for(int i=0;i<m;i++)
	{
		if(edge[i].u==b)
		edge[i].u=a;
		if(edge[i].v==b)
		edge[i].v=a;
	}
}
void kruskal()
{
	sort(edge+1,edge+1+m,cmp);
	for(int i=1;i<=m;i++)
	{
		if(edge[i].u!=edge[i].v)
		{
			//cout<<edge[i].u<<" "<<edge[i].v<<endl;
			if(i==strnum)
			{
				flag=1;
				return;
			}
			ans+=edge[i].w;
			k++;
			union_point(edge[i].u,edge[i].v);
		}
		if(k==n-1)
	    return;
	}
}

int main()
{
    int cas;
	cin>>cas;
	while(cas--)
	{
		flag=0;
		k=0;ans=0;
		cin>>n>>m>>strx>>stry; 
		for(int i=1;i<=m;i++)
		{
			int x,y,value;
			cin>>x>>y>>value;
			edge[i].u=x;
			edge[i].v=y;
			edge[i].w=value;
		}
		strnum=-1;
		for(int i=1;i<=m;i++)
		{
			if(edge[i].u==strx&&edge[i].v==stry)
			{
			
			strnum=i;
			break;
		    }
			if(edge[i].u==stry&&edge[i].v==strx)
			{
			
			strnum=i;
		    break;
		    }
		}
		kruskal();
		if(flag)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;	
}