Bellman-Ford||SPFA-POJ-3259-Wormholes

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 37628 Accepted: 13850
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 T 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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8
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.
Source

USACO 2006 December Gold

题目真的很难读懂，大致意思是路径是双向的，而虫洞是单向的，要求是否存在负环，用Bellman-Ford算法就可以了。

//
// main.cpp
// 最短路练习-F-Wormholes
//
// Created by 袁子涵 on 15/10/11.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//
// 532ms 800KB

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

#define INF 999999999

using namespace std;

typedef struct edge
{
    int S,E;
    long long int T;
}Edge;

int F,N,M,W;
Edge eg[10000];
long long int sum=0,dis[1000];

bool bellman_ford()
{
    bool flag=0;
    for (int i=1; i<N; i++) {
        dis[i]=INF;
    }
    dis[0]=0;
    for (int i=0; i<N; i++) {
        flag=0;
        for (int j=0; j<sum; j++) {
            if (dis[eg[j].E]>dis[eg[j].S]+eg[j].T && dis[eg[j].S]<INF) {
                dis[eg[j].E]=dis[eg[j].S]+eg[j].T;
                flag=1;
            }
        }
        if (flag==0) {
            break;
        }
    }
    for (int i=0; i<sum; i++) {
        if (dis[eg[i].E]>dis[eg[i].S]+eg[i].T && dis[eg[i].S]<INF) {
            dis[eg[i].E]=dis[eg[i].S]+eg[i].T;
            return 1;
        }
    }
    return 0;
}

int main(int argc, const char * argv[]) {
    cin >> F;
    int s,e,t;
    while (F--) {
        cin >> N >> M >> W;
        sum=0;
        for (int i=0; i<M; i++) {
            cin >> s >> e >> t;
            eg[sum].S=s-1;
            eg[sum].E=e-1;
            eg[sum++].T=t;
            eg[sum].S=e-1;
            eg[sum].E=s-1;
            eg[sum++].T=t;
        }
        for (int i=0; i<W; i++) {
            cin >> s >> e >> t;
            eg[sum].S=s-1;
            eg[sum].E=e-1;
            eg[sum++].T=-t;
        }
        if (bellman_ford()) {
            cout << "YES" << endl;
        }
        else
            cout << "NO" << endl;

    }
    return 0;
}

只是判断负环而已，所以还能用SPFA做。

//
// main.cpp
// POJ-3259-Wormholes
//
// Created by 袁子涵 on 15/11/28.
// Copyright © 2015年 袁子涵. All rights reserved.
//
// 125ms 812KB

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <queue>
#define MAXN 505
#define INF 0x3f3f3f3f
using namespace std;
int f,n,m,w;

struct Edge
{
    int v,cost;
    Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
void addedge(int u,int v,int w)
{
    E[u].push_back(Edge(v,w));
}
bool vis[MAXN];
int cnt[MAXN];
int dist[MAXN];
bool SPFA(int start)
{
    memset(vis, 0, sizeof(vis));
    for (int i=1; i<=n; i++) {
        dist[i]=INF;
    }
    dist[start]=0;
    vis[start]=1;
    queue<int>que;
    while (!que.empty()) {
        que.pop();
    }
    que.push(start);
    memset(cnt, 0, sizeof(cnt));
    cnt[start]=1;
    while (!que.empty()) {
        int u=que.front();
        que.pop();
        vis[u]=false;
        for (int i=0; i<E[u].size(); i++) {
            int v=E[u][i].v;
            if (dist[v]>dist[u]+E[u][i].cost) {
                dist[v]=dist[u]+E[u][i].cost;
                if (!vis[v]) {
                    vis[v]=1;
                    que.push(v);
                    if (++cnt[v]>n) {
                        return 0;
                    }
                }
            }
        }
    }
    return 1;
}
int main(int argc, const char * argv[]) {
    cin >> f;
    int a,b,c;
    while (f--) {
        cin >> n >> m >> w;
        for (int i=1; i<=n; i++) {
            E[i].clear();
        }
        for (int i=1; i<=m; i++) {
            scanf("%d%d%d",&a,&b,&c);
            addedge(a, b, c);
            addedge(b, a, c);
        }
        for (int i=1; i<=w; i++) {
            scanf("%d%d%d",&a,&b,&c);
            addedge(a, b, -c);
        }
        if(SPFA(1))
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}