Wormholes(SPFA+Bellman)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36860		Accepted: 13505

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

Sample Output

NO
YES
题解：
这道题意思是这个人的农田里有道路，还有虫洞，道路是双向的，虫洞单向，他喜欢虫洞旅行，想要从起点再回到起点时间在出发的时间之前，经过虫洞时间倒退；
做这道题其实就是判断最短路有没有负环的问题；因为只要有负环，就会无限循环，到起点自然就时间倒退了；
代码：SPFA

 1 #include<stdio.h>
 2 #include<queue>
 3 #include<string.h>
 4 using namespace std;
 5 const int INF=0x3f3f3f3f;
 6 const int MAXN=510;
 7 const int MAXM=10010*2;
 8 int dis[MAXN],vis[MAXN],used[MAXN],head[MAXM];
 9 int N,W,M,en,flot;
10 queue<int>dl;
11 struct Edge{
12     int from,to,value,next;
13 };
14 Edge edg[MAXM];
15 void initial(){
16     memset(dis,INF,sizeof(dis));
17     memset(vis,0,sizeof(vis));
18     memset(used,0,sizeof(used));
19     memset(head,-1,sizeof(head));
20     while(!dl.empty())dl.pop();
21     en=0;flot=0;
22 }
23 void print(){
24     if(flot)puts("YES");
25     else puts("NO");
26 }
27 void add(int u,int v,int w){
28     Edge E={u,v,w,head[u]};
29     edg[en]=E;
30     head[u]=en++;
31 }
32 void SPFA(int sx){
33     dis[sx]=0;vis[sx]=1;dl.push(sx);
34     used[sx]++;
35     while(!dl.empty()){
36         int k=dl.front();
37         dl.pop();
38         vis[k]=0;
39         if(used[k]>N){
40             flot=1;
41             break;
42                 }
43         for(int i=head[k];i!=-1;i=edg[i].next){
44             int v=edg[i].to;
45             if(dis[k]+edg[i].value<dis[v]){
46                 dis[v]=dis[k]+edg[i].value;
47                 if(!vis[v]){
48                     vis[v]=1;
49                     dl.push(v);
50                     used[v]++;
51                     if(used[v]>N){
52                         flot=1;return ;
53                     }
54                 }
55             }
56         }
57     }
58 }
59 void get(){
60     int F,a,b,c;
61     scanf("%d",&F);
62     while(F--){
63         initial();
64         scanf("%d%d%d",&N,&M,&W);
65         while(M--){
66             scanf("%d%d%d",&a,&b,&c);
67             add(a,b,c);
68             add(b,a,c);
69         }
70         while(W--){
71             scanf("%d%d%d",&a,&b,&c);
72             add(a,b,-c);
73         }
74         SPFA(1);
75         print();
76     }
77 }
78 int main(){
79     get();
80     return 0;
81 }

Bellman:

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int INF=0x3f3f3f3f;
 4 const int MAXN=510;
 5 const int MAXM=6000;
 6 int dis[MAXN];
 7 struct Edge{
 8     int u,v,w;
 9 };
10 Edge edg[MAXM];
11 int N,M,W,top;
12 bool Bellman(int sx){
13     int u,v,w;
14     memset(dis,INF,sizeof(dis));
15     dis[sx]=0;
16     for(int i=1;i<=N;i++){
17         for(int j=0;j<top;j++){
18             u=edg[j].u;v=edg[j].v;w=edg[j].w;
19             if(dis[u]+w<dis[v])dis[v]=dis[u]+w;
20         }
21     }
22     for(int i=0;i<top;i++){
23         u=edg[i].u;v=edg[i].v;w=edg[i].w;
24         if(dis[u]+w<dis[v])return false;
25     }
26     return true;
27 }
28 int main(){
29     int F;
30     int a,b,c;
31     scanf("%d",&F);
32     while(F--){
33         top=0;
34         scanf("%d%d%d",&N,&M,&W);
35         while(M--){
36             scanf("%d%d%d",&a,&b,&c);
37             edg[top].u=a;edg[top].v=b;edg[top++].w=c;
38             edg[top].u=b;edg[top].v=a;edg[top++].w=c;
39         }
40         while(W--){
41             scanf("%d%d%d",&a,&b,&c);
42             edg[top].u=a;edg[top].v=b;edg[top++].w=-c;
43         }
44         if(Bellman(1))puts("NO");
45         else puts("YES");
46     }
47     return 0;
48 }