| Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
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 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
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 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
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
struct node
{
int v, w, next;
}e[5500];
int k, n, m, w;
int head[5500];
bool vis[510];
int dis[510];
queue <int> q;
void adde(int u, int v, int w)
{
e[k].v = v;
e[k].w = w;
e[k].next = head[u];
head[u] = k++;
}
void input()
{
cin >> n >> m >> w;
int s, e, t;
k = 1;
memset(head, -1, sizeof(head));
for (int i = 0; i < m; i++)
{
cin >> s >> e >> t;
adde(s, e, t);
adde(e, s, t);
}
for (int i = 0; i < w; i++)
{
cin >> s >> e >> t;
adde(s, e, -t);
}
}
bool spfa(int x)
{
memset(vis, 0, sizeof(vis));
while (!q.empty())
q.pop();
memset(dis, 0x3f, sizeof(dis));
dis[x] = 0;
vis[x] = 1;
q.push(x);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = head[u]; i != -1; i = e[i].next)
{
int &v = e[i].v;
if (dis[v] > dis[u] + e[i].w)
{
dis[v] = dis[u] + e[i].w;
if (!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
if (dis[x] < 0)
return 1;
}
return 0;
}
void solve()
{
for (int i = 1; i <= n; i++)
{
if (spfa(i))
{
cout << "YES\n";
return;
}
}
cout << "NO\n";
}
int main()
{
int T;
cin >> T;
while (T--)
{
input();
solve();
}
return 0;
}
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