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.
题目意思:对于t组输入输出，有N个点，M条双向通路，K个虫洞。虫洞是一个单向的通路。问能否从某一个点出发，通过一些路径和虫洞，返回到出发点，并且时间还要早于出发时间，这样他就能够遇到他自己了。
解题思路：又是John，想都不用想又是USACO的题，美国果然是农业立国啊。在第二组样例中，John沿着1->2->3->1，从第一个点出发回到第一个点，所需时间为-1，这样他就能够遇到自己了。这该怎么思考？遇到虫洞会回到过去，时间回溯。实际上就是要求判断一个带有负权的有向网中是否存在负权值回路，
我们知道可以使用Bellman-Ford可以判断是否存在负环，因为出现了负值回路后会不断的松弛。这里我们用优化后的SPFA来实现。
原理：如果存在负权值回路，那么从源点到某一个点的最短路径可以无限缩短，某些顶点入队列将会超过N次（N为顶点个数）。
ps:代码有给出了另外一种判断是否有负环的方法，时间会更快。从任意一点出发求最短路，必定会求出到其他所有点的最短路，若其中某个点在一个负权回路中，那么它肯定会不断地走这个回路以使到这个点的花费越小越好，这时候该点到出发点的距离就会为负值，所以只需从一个点开始便能找到是否存在负回路，这里我们就只看从源点到源点的dis就可以了。
1 #include<cstdio>
2 #include<cstring>
3 #include<queue>
4 #include<algorithm>
5 #define INF 0x3f3f3f3f
6 #define maxs 10010
7 using namespace std;
8 struct node
9 {
10 int to;
11 int w;
12 } t;
13 vector<node>Map[maxs];
14 int dis[maxs];
15 int vis[maxs];
16 int n,m,k;
17 void add(int s,int e,int w)
18 {
19 t.to=e;
20 t.w=w;
21 Map[s].push_back(t);
22 }
23 void init()
24 {
25 int i;
26 memset(dis,INF,sizeof(dis));
27 memset(vis,0,sizeof(vis));
28 for(i=0; i<=n; i++)
29 {
30 Map[i].clear();
31 }
32 }
33 int SPFA(int s)
34 {
35 int cnt[maxs]= {0};///记录每个点入队次数
36 int i;
37 dis[s]=0;
38 vis[s]=1;
39 queue<int>q;
40 q.push(s);
41 cnt[s]++;
42 while(!q.empty())
43 {
44 int u=q.front();
45 /*if(cnt[u]>=n)
46 {
47 return 0;///这里用来判断入队列是否超过N次
48 }*/
49 q.pop();
50 vis[u]=0;
51 for(i=0; i<Map[u].size(); i++)
52 {
53 int to=Map[u][i].to;
54 if(dis[to]>dis[u]+Map[u][i].w)
55 {
56 dis[to]=dis[u]+Map[u][i].w;///松弛操作
57 if(dis[1]<0)///这种判断是否有负环的更快。如果起始点的dis[s]<0说明存在负环
58 {
59 return 0;
60 }
61 if(!vis[to])
62 {
63 vis[to]=1;
64 q.push(to);
65 cnt[to]++;
66 }
67 }
68 }
69 }
70 return 1;
71 }
72 int main()
73 {
74 int t,i,j,u,v,w;
75 scanf("%d",&t);
76 for(i=1; i<=t; i++)
77 {
78 scanf("%d%d%d",&n,&m,&k);
79 init();
80 while(m--)
81 {
82 scanf("%d%d%d",&u,&v,&w);
83 add(u,v,w);
84 add(v,u,w);
85 }
86 while(k--)
87 {
88 scanf("%d%d%d",&u,&v,&w);
89 add(u,v,-w);///虫洞时间取负值
90 }
91 if(SPFA(1))
92 {
93 printf("NO\n");
94 }
95 else
96 {
97 printf("YES\n");
98 }
99 }
100 return 0;
101 }
这里再附上使用邻接矩阵和Bellman-Ford算法的代码。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
struct Edge
{
int f;///起点
int t;///终点
int c;///距离
} edge[6000];
int dist[505];
int n,m,h,cnt;
int bellman_ford()
{
memset(dist,inf,sizeof(dist));
dist[1]=0;///起点
int i,j;
for(j=1; j<=n; j++)
{
for(i=1; i<=cnt; i++)
{
if(dist[edge[i].t]>dist[edge[i].f]+edge[i].c)///松弛操作
{
dist[edge[i].t]=dist[edge[i].f]+edge[i].c;
}
}
}
int flag=1;
for(i=1; i<=cnt; i++)///遍历所有的边
{
if(dist[edge[i].t]>dist[edge[i].f]+edge[i].c)
{
flag=0;///存在从源点可达的负权回路
break;
}
}
return flag;
}
int main()
{
int i,t;
scanf("%d",&t);
while(t--)
{
cnt=0;
scanf("%d %d %d",&n,&m,&h);
int u,v,w;
for(i=1; i<=m; i++)
{
cnt++;
scanf("%d %d %d",&u,&v,&w);
edge[cnt].f=u;
edge[cnt].t=v;
edge[cnt].c=w;
cnt++;
edge[cnt].f=v;
edge[cnt].t=u;
edge[cnt].c=w;
}
for(i=1; i<=h; i++)
{
cnt++;
scanf("%d %d %d",&u,&v,&w);
edge[cnt].f=u;
edge[cnt].t=v;
edge[cnt].c=-w;
}
if(bellman_ford())
printf("NO\n");
else
printf("YES\n");
}
return 0;
}
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