Wormholes POJ - 3259 SPFA 最短路

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler backT seconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

第二发spfa -。-至于为啥要用邻接表主要是这个有重边无向的path可能会被有向的wormhole覆盖又不是求最短路所以每条边必须保存

所以邻接表是最好的选择因为直接上一波vector就好了嘛

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
#define clr(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
int f,n,m,w,visit[505],index[505],dis[505];
struct node
{
	int to;
	int power;
};
vector <node> ma[505];//使用邻接表来存数据 ma[i][j].to i是出发点 j是边的编号 to是到达点 上一次直接用简单粗暴的结构体二维数组存的图 贼粗暴 

int spfa()
{
	queue<int> q;
	q.push(1);//始点是一的单源最短路 
	dis[1]=0;
	visit[1]=1;
	index[1]=1;
	int in=0;
	while(!q.empty())
	{
		in=q.front();
		q.pop();
		visit[in]=0;
		if(index[in]>n)
		{
			return 1;
		}
		for(int i=0;i<ma[in].size();i++)
		{
			int to=ma[in][i].to;
			if(dis[to]>dis[in]+ma[in][i].power)
			{
		        dis[to]=dis[in]+ma[in][i].power;
				if(!visit[to])
				{
					visit[to]=1;
					q.push(to);
					index[to]++;						
				}
			}
		}
	}
	return 0;
}

int main()
{
	ios::sync_with_stdio(false);
	cin>>f;
	while(f--)
	{
		cin>>n>>m>>w;
		for(int i=0;i<=504;i++)
		ma[i].clear();
		clr(visit,0)
		clr(index,0)
		clr(dis,inf)
		for(int i=1;i<=m;i++)
		{
			int s,e,p;
			node g;
			cin>>s>>e>>p;
			g.to=e;
			g.power=p;
			ma[s].push_back(g);
			g.to=s;
			ma[e].push_back(g);
			
		}
		for(int i=1;i<=w;i++)
		{
			int s,e,p;
			node g;
			cin>>s>>e>>p;
			g.to=e;
			g.power=-p;
			ma[s].push_back(g);
		}
		if(spfa()) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0; 
}

Wormholes POJ - 3259 SPFA 最短路

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