Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 45420 | Accepted: 16767 |
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.
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
题意：
有多组数据，第一个数代表有多少组数据。对于每一组数据，第一行有三个数，代表有n个点，m条双向边， k条单向边。
接下来有m行代表a与b之间有一条权值为d的边，接下来k行，代表从a到b有一条权值为 – d的单向边。
注意，这里需要注意，题目中给的单向边的权值是正的，但处理时要将它变为负的。
然后询问你是否会回到起始点。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int n, m, w;
int first[233333], next[233333], d[233333], tot = 0;
bool used[233333];
struct qer
{
int from, to, cost;
}es[233333];
queue <int> q;
void init()
{
memset(first, -1, sizeof(first));
memset(d, 0x3f, sizeof(d));
memset(used, 0, sizeof(used));
tot = 0;
}
void build(int f, int t, int d)
{
es[++tot] = (qer){f, t, d};
next[tot] = first[f];
first[f] = tot;
}
bool spfa(int s)
{
d[s] = 0;
used[s] = 1;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
used[u] = 0;
if(d[s] < 0)
return 0;
for(int i = first[u];i != -1;i = next[i])
{
int v = es[i].to;
if(d[v] > d[u] + es[i].cost)
{
d[v] = d[u] + es[i].cost;
if(!used[v])
{
q.push(v);
used[v] = 1;
}
}
}
}
return 1;
}
int main()
{
int f;
cin>>f;
while(f--)
{
init();
scanf("%d%d%d", &n, &m, &w);
for(int i = 1;i <= m;i ++)
{
int f, t, d;
scanf("%d%d%d", &f, &t, &d);
build(f, t, d);
build(t, f, d);
}
for(int i = 1;i <= w;i ++)
{
int f, t, d;
scanf("%d%d%d", &f, &t, &d);
build(f, t, -d);
}
bool ans = spfa(1);
if(ans == 1)
puts("NO");
else
puts("YES");
}
return 0;
}