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.
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <iomanip> #include <deque> #include <bitset> //#include <unordered_set> //#include <unordered_map> //#include <bits/stdc++.h> //#include <xfunctional> #define ll long long #define PII pair<int, int> #define rep(i,a,b) for(ll i=a;i<b;i++) #define dec(i,a,b) for(ll i=a;i>=b;i--) #define pb push_back #define mp make_pair using namespace std; int dir[5][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } ,{ 0,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; const double pi = 3.14159265358979; const int mod = 1e9 + 7; const int N = 2500 * 2 + 200 + 10; //if(x<0 || x>=r || y<0 || y>=c) inline ll read() { ll x = 0; bool f = true; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); } while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return f ? x : -x; } int n, m, w; struct edge { int u, v; int t; }; edge e[N]; bool bf() { vector<int> d(510, 10000); d[1] = 0; for (int i = 1; i < n; i++) { int fg = 1; for (int j = 0; j < 2 * m + w; j++) { int u = e[j].u; int v = e[j].v; int t = e[j].t; if (d[v] > d[u] + t) { d[v] = d[u] + t; fg = 0; } } if (fg) return false; } for(int i=0;i<2 * m + w; i++) { if (d[e[i].v] > d[e[i].u] + e[i].t) return true; } return false; } int main() { int T; cin >> T; while (T--) { cin >> n >> m >> w; int u, v, t; int tmp = 0; for (int i = 1; i <= m; i++) { cin >> u >> v >> t; e[tmp].u = u; e[tmp].v = v; e[tmp++].t = t; e[tmp].u = v; e[tmp].v = u; e[tmp++].t = t; } for (int i = 1; i <= w; i++) { cin >> u >> v >> t; e[tmp].u = u; e[tmp].v = v; e[tmp++].t = -t; } if (bf()) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }