Wormholes(POJ 3259)

Wormholes
Time Limit:2000MS Memory Limit:65536KB 64bit
IO Format:%I64d & %I64u

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 22 3 11 3 43 1 3
2 3 43 2 1
1 2 3
3 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

题意：John在农场里探索，发现每个农场包括n块田地，m条双向路径以及w个单向虫洞。他想知道他从任意一块田地出发，能否再回到出发点。
分析：可以将题抽象成有n个顶点，m条双向边（权值为t）和w条单向边（权值为-t）。判断是否存在负环。用Bellman-Ford算法判断是否存在负环，如果不存在，则迭代n次后可求得其最短路；否则，最短路不存在（沿着负环一直走，路会更短）

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;
const int MAXN = 500 + 10;
const int MAXM = 5000 + 1000;
const int INF = 0x3f3f3f3f;
int n,m,w,k;
int f[MAXN];
//从顶点s指向顶点e的权值为t的边
struct farm
{
int s,e,t;
}map[MAXM];
//记录一条从顶点x到顶点y权值为z的路径
void join(int x,int y,int z)
{
map[k].s = x;
map[k].e = y;
map[k].t = z;
k++;
}
//如果返回1则存在负圈，否则不存在
int Bellman_Ford()
{
memset(f,INF,sizeof(f));
f[1] = 0;
//求最短路
for(int i = 0;i < n;i++)
{
for(int j = 0;j < k;j++)
{
int u = map[j].s;
int v = map[j].e;
int w = map[j].t;
if(f[u] + w < f[v])
f[v] = f[u] + w;
}
}
for(int i = 0;i < n;i++)
{
for(int j = 0;j < k;j++)
{
int u = map[j].s;
int v = map[j].e;
int w = map[j].t;
if(f[u] + w < f[v])   //若还能更新，则存在负圈
return 1;
}
}
return 0;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&n,&m,&w);
int x,y,z;
k = 0;
for(int i = 0;i < m;i++)
{
scanf("%d %d %d",&x,&y,&z);
join(x,y,z);
join(y,x,z);
}
for(int i = 0;i < w;i++)
{
scanf("%d %d %d",&x,&y,&z);
join(x,y,-z);
}
if(Bellman_Ford())
printf("YES\n");
else printf("NO\n");
}
return 0;
}