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

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.

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 1000000
int 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;
}