标签：3259 int edge farm 无向 poj time FJ 负权
http://poj.org/problem?id=3259
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
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.
题目大意：
时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。
解题思路：
裸的负权最短路问题，SPAF Bellman-Ford解决。
1 #include<iostream>
2 #include<cstdio>
3 using namespace std;
4 #define INF 0x3f3f3f3f
5 #define N 10100
6 int nodenum, edgenum, w, original=1; //点，边，起点
7
8 typedef struct Edge //边
9 {
10 int u;
11 int v;
12 int cost;
13 }Edge;//边的数据结构
14
15 Edge edge[N];//边
16
17 int dis[N];//距离
18
19 bool Bellman_Ford()
20 {
21 for(int i = 1; i <= nodenum; ++i) //初始化
22 dis[i] = (i == original ? 0 : INF);
23 int F=0;
24 for(int i = 1; i <= nodenum - 1; ++i)//进行nodenum-1次的松弛遍历
25 {
26 for(int j = 1; j <= edgenum*2+w; ++j)
27 {
28 if(dis[edge[j].v] > dis[edge[j].u] + edge[j].cost) //松弛（顺序一定不能反~）
29 {
30 dis[edge[j].v] = dis[edge[j].u] + edge[j].cost;
31 F=1;
32 }
33 }
34 if(!F)
35 break;
36 }
37 //与迪杰斯特拉算法类似，但不是贪心！
38 //并没有标记数组
39 //本来松弛已经结束了
40 //但是因为由于负权环的无限松弛性
41 bool flag = 1; //判断是否含有负权回路
42 //如果存在负权环的话一定能够继续松弛
43 for(int i = 1; i <= edgenum*2+w; ++i)
44 {
45 if(dis[edge[i].v] > dis[edge[i].u] + edge[i].cost)
46 {
47 flag = 0;
48 break;
49 }
50 }
51 //只有在负权环中才能再松弛下去
52 return flag;
53 }
54
55 int main()
56 {
57 int t;
58 scanf("%d",&t);
59 while(t--)
60 {
61
62 scanf("%d %d %d", &nodenum, &edgenum, &w);
63
64 for(int i = 1; i <= 2*edgenum; i+=2)//加上道路，双向边
65 {
66 scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
67 edge[i+1].u=edge[i].v;
68 edge[i+1].v=edge[i].u;
69 edge[i+1].cost=edge[i].cost;
70 }
71 for(int i =2*edgenum+1; i <= 2*edgenum+w; i++)//加上虫洞，单向边，负权
72 {
73 scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
74 edge[i].cost=-edge[i].cost;
75 }
76 if(Bellman_Ford())//没有负环
77 printf("NO\n");
78 else
79 printf("YES\n");
80 }
81 return 0;
82 }
标签：3259,int,edge,farm,无向,poj,time,FJ,负权 来源： https://www.cnblogs.com/jiamian/p/11271971.html
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