带负权的最短路bellman_ford——POJ 3259 Wormholes题解

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

解题思路

正权最短路用dij是使用的数组，这里因为是带负权的，所以使用结构体，自定义一个。然后以每个节点为根节点遍历，然后判断是否有负权。犯低级错误，数组下标边界问题没搞对了，错了n发

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define inf 1000000int n, m1, m2,flag = 0;int num = 0, dis[10010];struct node{int x,y,cnt;}road[10000];int bellman_ford(){memset(dis, inf, sizeof(dis));dis[road[0].x] = 0;for(int i = 1; i < n; i++){for(int j = 0; j < num; j++){if(dis[road[j].y] > dis[road[j].x] + road[j].cnt)dis[road[j].y] = dis[road[j].x] + road[j].cnt;}}for(int i = 0; i < num ; i++)if(dis[road[i].y] > dis[road[i].x] + road[i].cnt)flag = 1;return flag;}int main(){int t;scanf("%d", &t);while(t--){flag = 0;num = 0;memset(road,0,sizeof(road));scanf("%d%d%d", &n, &m1, &m2);for(int i = 0; i < m1; i++){scanf("%d%d%d", &road[num].x, &road[num].y, &road[num].cnt);num++;road[num].x = road[num-1].y;road[num].y = road[num-1].x;road[num].cnt = road[num-1].cnt;num++; }for(int i = 0; i < m2; i++){scanf("%d%d%d", &road[num].x, &road[num].y, &road[num].cnt);road[num].cnt = -road[num].cnt;num++;}if(!bellman_ford())printf("NO\n");else printf("YES\n");}return 0;}