使用SPFA算法解决带权重循环问题-优快云博客

本文探讨了如何利用SPFA算法解决具有负权重循环的路径问题，包括图中可能存在的重边和自环情况。通过实例POJ 3259 Wormholes，详细解释了解决方案和算法应用。

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负权回路判断，有重边有自环，万能的spfa解决

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39617		Accepted: 14559

Description

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 back Tseconds.

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.

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#define Min(a,b) a>b?b:a;
#define Max(a,b) a>b?a:b;
using namespace std;
const int SIZE=5e2+10;
const int maxn=1<<30;
struct node{
    int v;int w;
    node(){}
    node(int vv,int ww):v(vv),w(ww){}
};
vector<vector<node> >G;
int cost[SIZE][SIZE];
int dis[SIZE];
bool spfa(int s,int n){
    int uptimes[SIZE];
    memset(uptimes,0,sizeof(uptimes));
    for(int i=1;i<=n;i++)
        dis[i]=maxn;
    dis[s]=0;
    queue<int>q;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=0;i<G[u].size();i++){
            int v=G[u][i].v;
            int w=G[u][i].w;
            if(w!=cost[u][v])continue;
            if(dis[v]<=dis[u]+w)continue;
            dis[v]=dis[u]+w;
            q.push(v);
            uptimes[v]++;
            if(uptimes[v]>=n)return true;
        }
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        memset(cost,0,sizeof(cost));
        int n,m,x,u,v,w;
        scanf("%d%d%d",&n,&m,&x);
        G.clear();
        G.resize(n+10);
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&w);
            if(u==v)continue;
            G[u].push_back(node(v,w));
            G[v].push_back(node(u,w));
            if(cost[u][v]){
                cost[u][v]=cost[v][u]=Min(cost[v][u],w);
            }
            else cost[u][v]=cost[v][u]=w;
        }
        for(int i=0;i<x;i++){
            scanf("%d%d%d",&u,&v,&w);
            G[u].push_back(node(v,-w));
            cost[u][v]=Min(cost[u][v],-w);
        }
        if(spfa(1,n))printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}