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POJ 3259 Wormholes （图论---最短路 Bellman-Ford || SPFA）

2024-07-06 14:57:46   230人阅读

链接：http://poj.org/problem?id=3259

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.

题目大意：

有一个农场，里面有一些奇怪的单向虫洞，通过虫洞的入口到出口可以让时间倒退，农场主想知道他是否能够通过农场中的路和虫洞走回出发点，并且使时间倒退，给出农场中的N块地， M条路以及W个虫洞；

分析：

由题意可以看出实际目标位寻找负值回路，所以可以使用Bellman - Ford或者SPFA判断图中是否存在负权回路即可；

代码：

（Bellman - Ford）:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define MAXN 505
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;

typedef struct Edge_
{
    int u, v, t;
}Edge;

Edge edge[6005];
int res[MAXN];   //存储源点到每个顶点的最短距离值；
int cas, n, m, w, s, e, t;

void READ_DATA()
{
    scanf("%d %d %d", &n, &m, &w);
    for(int i=0; i<m; i++) {
        scanf("%d %d %d", &s, &e, &t);
        edge[i].u = s, edge[i].v = e, edge[i].t = t;
        edge[i+m].u = e, edge[i+m].v = s, edge[i+m].t = t;
    }
    for(int i=0; i<w; i++) {
        scanf("%d %d %d", &s, &e, &t);
        edge[i+2*m].u = s, edge[i+2*m].v = e, edge[i+2*m].t = -t;
    }
}

bool Bellman(int n, int m, int src)   //判断是否有负环；
{
    memset(res, 0x1f, sizeof(res));  //初始化每个顶点的最短距离为无穷大；
    res[src] = 0;                 //源点的距离为0；
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(res[edge[j].u] + edge[j].t < res[edge[j].v]) res[edge[j].v] = res[edge[j].u] + edge[j].t;
        }
    }
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(res[edge[j].u] + edge[j].t < res[edge[j].v]) return true;
        }
    }
    return false;
}

int main()
{
    scanf("%d", &cas);
    while(cas--) {
        READ_DATA();
        if(Bellman(n, 2*m+w, 1)) puts("YES");
        else puts("NO");
    }
    return 0;
}

（SPFA）:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define MAXN 505
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;

int res[MAXN];   //存储源点到每个顶点的最短距离值；
int map[MAXN][MAXN];
int cnt[MAXN];   //每个点入队次数；
int que[MAXN*MAXN];   //队列；
bool In_que[MAXN];    //标记一个点是否出现在队列中；
int front, rear, cas;  //队首，队尾，测试数据个数；
int n, m, w, s, e, t;

void READ_DATA()
{
    scanf("%d %d %d", &n, &m, &w);
    memset(map, 0x1f, sizeof(map));
    for(int i=0; i<m; i++) {
        scanf("%d %d %d", &s, &e, &t);
        map[s][e] = map[e][s] = map[s][e] > t ? t : map[s][e];
    }
    for(int i=0; i<w; i++) {
        scanf("%d %d %d", &s, &e, &t);
        map[s][e] = map[s][e] < -t ? map[s][e] : -t;
    }
}

bool SPFA(int n, int src)
{
    RST(cnt), RST(In_que);
    front = rear = 0;
    que[++rear] = src;
    cnt[src]++;
    memset(res, 0x1f, sizeof(res));
    res[src] = 0;
    while(front < rear) {
        int current = que[++front];
        In_que[current] = 0;
        for(int i=1; i<=n; i++) {
            if(res[current] + map[current][i] < res[i]) {
                res[i] = res[current] + map[current][i];
                if(!In_que[i]) {
                    que[++rear] = i, cnt[i]++;
                    if(cnt[i] >= n) return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    scanf("%d", &cas);
    while(cas--) {
        READ_DATA();
        if(SPFA(n, 1)) puts("YES");
        else puts("NO");
    }
    return 0;
}

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