Wormholes
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 46024 | Accepted: 17006 |
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 toF (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 toE that also moves the traveler backT 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.
题意：John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。 现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。
虫洞问题：给n，m，W，n是有n个点，m是有m个正权，w是有w个负权。
题解：用到Bellman--Ford来解决负权值，思想是将所有边都“松弛”，用到关键的
for(i=1; i<m; i++)//m是所有边数。
{
if(div[ v[i] ] > div[ u[i] ] + w[i])
div[ v[i] ] = div[ u[i] ] + w[i];
}
来对各个边进行压缩松弛；
不断循环，当松弛到点数减一时就压缩完了，如果还能压缩说明存在负权值来影响总的压缩。
<span style="font-family:Arial;">#include <iostream>
using namespace std;
const int inf=0x3f3f3f3f;
struct node
{
int u, v, w;
} Map[10100];
int n, x, y, z;
int div[10100];
int Bellman(int m)
{
int i, j, check;
for(i = 1; i <= n-1; i++)
{
check = 0;
for(j = 1; j <= m ; j++)
{
if(div[Map[j].u] > div[Map[j].v] + Map[j].w)
{
div[Map[j].u] = div[Map[j].v] + Map[j].w;
check = 1;
}
}
if(check == 0)
break;
}
int flag = 0;
for(i = 1; i <= m; i++)
{
if(div[Map[i].u] > div[Map[i].v] + Map[i].w)
{
flag = 1;
break;
}
}
if(flag == 1)
return 0;
else
return 1;
}
int main()
{
int k, i, j, m, W;
cin>>k;
while(k--)
{
cin>>n>>m>>W;
for(j =0, i = 1; i <= m; i++)
{
cin>>x>>y>>z;
j++;
Map[j].u = x;
Map[j].v = y;
Map[j].w = z;
j++;
Map[j].u = y;
Map[j].v = x;
Map[j].w = z;
}
for(i = 1; i <= W; i++)
{
cin>>x>>y>>z;
j++;
Map[j].u = x;
Map[j].v = y;
Map[j].w = -z;
}
for(i = 1; i <= n; i++)
{
div[i] = inf;
}
div[1] = 0;
if(Bellman(j))
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
return 0;
}
</span>
