POJ 3259 Wormholes(BellmanFord)
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
题目链接
BellmanFord算法模板题，最短路径算法—Bellman-Ford(贝尔曼-福特)算法分析与实现(C/C++)
AC代码:
#pragma comment(linker, "/STACK:102400000,102400000")
//#include <bits/stdc++.h>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <climits>
#include <stdlib.h>
#include <deque>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <string>
#include <vector>
#include <bitset>
#include <complex>
#include <functional>
#include <fstream>
#include <ctime>
#include <stdexcept>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int maxn = 5e2+5;
const int mod = 1e6;
const double eps = 1e-8;
const double pi = asin(1.0)*2;
//const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}
struct Edge {
int u;
int v;
int cost;
Edge(int _u = 0, int _v = 0, int _cost = 0): u(_u), v(_v), cost(_cost) {}
};
int T;
int n, m, w;
int s, e, t;
int dist[maxn];
vector<Edge> E;
bool BellmanFord() {
mem(dist, INF);
dist[1]= 0;
for (int i = 1; i < n; ++i) {
bool flag = 0;
for (int j = 0; j < int(E.size()); ++j) {
int u = E[j].u;
int v = E[j].v;
int cost = E[j].cost;
if (dist[v] > dist[u] + cost) {
dist[v] = dist[u] + cost;
flag = 1;
}
}
if (!flag) {
return 1;
}
}
for (int j = 0; j < int(E.size()); ++j) {
if (dist[E[j].v] > dist[E[j].u] + E[j].cost) {
return 0;
}
}
return 1;
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
read(T);
while (T--) {
read(n); read(m); read(w);
E.clear();
for (int i = 1; i <= m; ++i) {
read(s); read(e); read(t);
Edge add;
add.u = s;
add.v = e;
add.cost = t;
E.pb(add);
add.u = e;
add.v = s;
E.pb(add);
}
for (int i = 1; i <= w; ++i) {
read(s); read(e); read(t);
Edge add;
add.u = s;
add.v = e;
add.cost = -t;
E.pb(add);
}
if (BellmanFord()) {
printf("NO\n");
}
else {
printf("YES\n");
}
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}
