POJ3259 Wormholes(Bellman-ford 负环)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 33194		Accepted: 12085

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:

还是运用bellman-ford算法，求负环。

代码如下:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 10000;  //开了5002 RE，索性爆炸点开个10000.

#define inf 100000000

int all; //表示边数
int n,m,w; //分别表示顶点数，邻接边，虫洞个数
int cnt;

struct point
{
int start;
int end;
int seconds;
} a[maxn];

int dis[maxn];

bool bellman()
{
int i,j;
bool flag;
memset(dis,inf,sizeof(dis));
dis[0] = 0;
for(i=1; i<n; i++)  //因为0这个点已经初始化，所以从1这个点开始
{
flag = false;
for(j=0; j< all; j++)   //all为总的边数，包含虫洞
{
if(dis[a[j].start] > dis[a[j].end] + a[j].seconds)
{
dis[a[j].start] = dis[a[j].end] + a[j].seconds;
flag = true;
}
}
if(!flag)
{
break;
}
}
for(j=0; j<all; j++)
{
if(dis[a[j].start] > dis[a[j].end] + a[j].seconds)
{
dis[a[j].start] = dis[a[j].end] + a[j].seconds;
return true;  //存在负环
}
}
return false; //不存在负环，也就是回不去
}

int main()
{
int F,i;
int num1,num2,seconds;
//freopen("111","r",stdin);
cin>>F;
while(F--)
{
all = 0;
cin>>n>>m>>w;
for(i=0; i<m; i++) //普通的道路
{
cin>>num1>>num2>>seconds;
a[all].start = num1;
a[all].end = num2;
a[all++].seconds = seconds;
a[all].start = num2;
a[all].end = num1;
a[all++].seconds = seconds;
}
for(i=0; i<w; i++)
{
cin>>num1>>num2>>seconds;
a[all].start = num1;
a[all].end = num2;
a[all++].seconds = -seconds;
}
if(bellman())
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
return 0;
}