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Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 52515 Accepted: 19552

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.

题意 : 是一个人要穿越到未来，但是之后还要回去，并且回去的时间要在他穿越之前。输入m组无向边，输入w组有向边。在构成最短路之后查找是否有负环，如果有则农民可以回来即YES，没有为NO。注意回来的有向边因为是时间倒流，花的时间是负值，这样才能构成负环。

解题思路：由题意知（在虫洞中）往回走（单向边）的时候的边权是负数，针对负权边，（万能方法）我们用Bellman-ford，数据小的时候，可直接用floyed。

代码如下：

（Bellman-ford）

#include<stdio.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int n,m,W,dis[25505];
int f,t;
struct p
{
int u,v,w;
} a[25505];
void ford()
{
for(int i=1; i<=n; i++)
dis[i]=inf;//初始化dis数组1号顶点到其他顶点的距离

//算法核心代码
for(int i=1; i<=n-1; i++)
for(int j=1; j<=f; j++)
if(dis[a[j].v]>dis[a[j].u]+a[j].w)
dis[a[j].v]=dis[a[j].u]+a[j].w;
int flag=0;
for(int i=1; i<=f; i++)//查找负权边
{
if(dis[a[i].v]>dis[a[i].u]+a[i].w)//之前已经松弛了n-1次，若在可以松弛，则存在负权回路
{
flag=1;
break;
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
int main()
{
scanf("%d",&t);
while(t--)
{
int i,b,c,d;
f=1;
scanf("%d %d %d",&n,&m,&W);//输入路径数目，区域数目
for( i=1; i<=m; i++)
{
scanf("%d %d %d",&b,&c,&d);
a[f].u=b;//双向路径
a[f].v=c;
a[f++].w=d;
a[f].u=c;
a[f].v=b;
a[f++].w=d;
}
for(i=m+1; i<=W+m; i++)//虫洞单向
{
scanf("%d %d %d",&b,&c,&d);
a[f].u=b;
a[f].v=c;
a[f++].w=-d;//负权边
}
ford();
}
return 0;
}

（floyd）#include<stdio.h>
#include<string.h>
int N,M,W,u,v,w,e[505][505];
int floyd()
{
for(int k=1; k<=N; k++)
for(int i=1; i<=N; i++)
{
for(int j=1; j<=N; j++)
{
if(e[i][j]>e[i][k]+e[k][j])
e[i][j]=e[i][k]+e[k][j];
}
if(e[i][i]<0)
return 1;
}
return 0;

}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&N,&M,&W);
memset(e,0x3f3f3f3f,sizeof(e));
for(int i=1; i<=N; i++)//距离数组初始化
e[i][i]=0;
for(int i=1; i<=M; i++)
{
scanf("%d %d %d",&u,&v,&w);//输入边权
if(e[u][v]>w)
e[u][v]=e[v][u]=w;
}
for(int i=1; i<=W; i++)
{
scanf("%d %d %d",&u,&v,&w);
e[u][v]=-w;
}
if(!floyd())
printf("NO\n");
else
printf("YES\n");

}
return 0;
}

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