| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 28499 | Accepted: 10302 |
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
题意：John的农场里field块地，path条路连接两块地，hole个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
思路：bellman_ford。由于存在负权边，Dijkstra便不能用了。题目简化下，就是看图中有没有负权环，有的话John可以无限次走这个环，使得时间一定能得到一个负值。所以有的存在负环话就是可以，没有的话就是不可以了。
#include <iostream>
#include <cstdio>
using namespace
std;
#define fMax 505
#define eMax 5205
#define wMax 99999
struct A
{
int sta,end,time;
}edge[eMax];
int point_num,edge_num,dict[fMax];
bool bellman_ford()
{
for(int
i=2;i<=point_num;i++)
dict[i]=wMax;
for(int i=1;i<=point_num;i++)
{
bool
fals=0;
for(int
j=1;j<=edge_num;j++)
{
int
v=edge[j].sta;
int
u=edge[j].end;
int
w=edge[j].time;
if(dict[v]>dict[u]+w)
{
dict[v]=dict[u]+w;
fals=1;
}
}
if(!fals)
break;
}
for(int
j=1;j<=edge_num;j++)
{
int
v=edge[j].sta;
int
u=edge[j].end;
int
w=edge[j].time;
if(dict[v]>dict[u]+w)
{
dict[v]=dict[u]+w;
return 0;
}
}
return 1;
}
int main()
{
int farm;
scanf("%d",&farm);
while(farm--)
{
int
field,path,hole;
scanf("%d %d
%d",&field,&path,&hole);
int s,e,t,k=0;
for(int
i=1;i<=path;i++)
{
scanf("%d %d
%d",&s,&e,&t);
edge[++k].sta=s;
edge[k].end=e;
edge[k++].time=t;
edge[k].sta=e;
edge[k].end=s;
edge[k].time=t;
}
for(int
i=1;i<=hole;i++)
{
scanf("%d %d
%d",&s,&e,&t);
edge[++k].sta=s;
edge[k].end=e;
edge[k].time=-t;
}
point_num=field;
edge_num=k;
if(!bellman_ford())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}