Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 36717 | Accepted: 13438 |
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.
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.
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.
Source
70ms
#include<iostream> //79ms
#include<cstdio>
#include<cstring>
#include<cmath>
#define INF 10000000
using namespace std;
struct node
{
int u,v,w;
} edge[5500];
int low[5500];
int n,m,z;
int num=0;
int Bellman()
{
for(int i=0; i<=n; i++)
low[i]=INF;
for(int i=0; i<n-1; i++)
{
int flag=0;
for(int j=0; j<num; j++)
{
if(low[edge[j].u]+edge[j].w<low[edge[j].v])
{
low[edge[j].v]=low[edge[j].u]+edge[j].w;
flag=1;
}
}
if(flag==0) //存在负权回路
break;
}
for(int j=0; j<num; j++) //推断负权回路
{
if(low[edge[j].u]+edge[j].w<low[edge[j].v])
return 1;
}
return 0;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&z);
int a,b,c;
num=0;
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=c;
edge[num].u=b;
edge[num].v=a;
edge[num++].w=c;
}
for(int i=1; i<=z; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=-c;
}
if(Bellman())
printf("YES
");
else
printf("NO
");
}
}
700ms
#include<iostream> //挨个点遍历
#include<cstdio>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
int u,v,w;
} edge[5500];
int low[550];
int n,m,z;
int num=0;
int Bellman(int u0)
{
for(int i=0; i<=n; i++)
low[i]=INF;
low[u0]=0;
for(int i=0; i<n; i++) //递推n次,让其构成环来推断
{
int flag=0;
for(int j=0; j<num; j++)
{
if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])
{
low[edge[j].v]=low[edge[j].u]+edge[j].w;
flag=1;
}
}
if(flag==0) //存在负权回路(降低时间)
break;
}
if(low[u0]<0)
return 1;
return 0;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&z);
int a,b,c;
num=0;
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=c;
edge[num].u=b;
edge[num].v=a;
edge[num++].w=c;
}
for(int i=1; i<=z; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=-c;
}
int biao=1;
for(int i=1; i<=n; i++)
{
if(Bellman(i))
{
printf("YES
");
biao=0;
break;
}
}
if(biao)
printf("NO
");
}
}
/*
780ms
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
int u,v,w;
} edge[5500];
int low[5500];
int n,m,z;
int num=0;
int Bellman(int u0)
{
for(int i=0; i<=n; i++)
low[i]=INF;
low[u0]=0; //初始化
for(int i=0; i<n-1; i++) //n-1次
{
int flag=0;
for(int j=0; j<num; j++)
{
if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v]) //不同点
{ //存在low[edge[j].u]!=INF,就必须有low[u0]=0;初始化
low[edge[j].v]=low[edge[j].u]+edge[j].w;
flag=1;
}
}
if(flag==0) //存在负权回路
break;
}
for(int j=0; j<num; j++)
{
if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])
return 1;
}
return 0;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&z);
int a,b,c;
num=0;
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=c;
edge[num].u=b;
edge[num].v=a;
edge[num++].w=c;
}
for(int i=1; i<=z; i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[num].u=a;
edge[num].v=b;
edge[num++].w=-c;
}
int biao=1;
for(int i=1; i<=n; i++)
{
if(Bellman(i))
{
printf("YES
");
biao=0;
break;
}
}
if(biao)
printf("NO
");
}
}
*/