Wormholes

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

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.

题意：

有n个点，m条普通的双向的路，w条单向的虫洞，问是否能从某个点出发走一些路再回来时回到过去。

即判断是否有负环。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<queue>
 5 #include<stack>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<vector>
 9 #define N 50000//记得开大一点，做的时候开5000RE了好几遍
10 #define inf 0x3f3f3f
11 using namespace std;
12 
13 struct node
14 {
15     int to,w,next;
16 } s[N];
17 
18 int n;
19 int tot;
20 int vis[N];
21 int dis[N];
22 int head[N];
23 int num[N];
24 
25 void add(int u,int v,int w)//记得这个函数存的是边的信息（编号为tot的边的指向以及权值），而不是点的信息，
26 {
27     s[tot].to = v;
28     s[tot].w = w;
29     s[tot].next = head[u];
30     head[u] = tot++;//边的个数
31 }
32 
33 int spfa()
34 {
35     queue<int> q;
36     memset(num,0,sizeof(num));
37     memset(vis,0,sizeof(vis));
38     for(int i = 1; i <= n; i++)//这里初始化可以用memset
39     {
40         dis[i] = inf;
41     }
42     dis[1] = 0;
43     num[1] = 1;
44     vis[1] = 1;
45     q.push(1);
46     while(q.size())
47     {
48         int u = q.front();
49         q.pop();
50         vis[u] = 0;
51         for(int i = head[u]; i != -1; i = s[i].next)
52         {
53             int v = s[i].to;
54             if(dis[v] > dis[u] + s[i].w)
55             {
56                 dis[v] = dis[u] + s[i].w;
57                 if(vis[v] == 0)
58                 {
59                     num[v]++;
60                     if(num[v] >= n)//如果入队次数大于点的个数则存在负环
61                         return 1;
62                     vis[v] = 1;
63                     q.push(v);
64                 }
65             }
66         }
67     }
68     return 0;
69 }
70 
71 int main()
72 {
73     int t,x,y;
74     int u,v,w;
75     scanf("%d",&t);
76     while(t--)
77     {
78         tot = 1;//tot为0或1没差
79         memset(head,-1,sizeof(head));
80         scanf("%d %d %d",&n,&x,&y);
81         while(x--)
82         {
83             scanf("%d %d %d",&u,&v,&w);
84             add(u,v,w);
85             add(v,u,w);
86         }
87         while(y--)
88         {
89             scanf("%d %d %d",&u,&v,&w);
90             add(u,v,w*(-1));
91         }
92         if(spfa())
93             printf("YES\n");
94         else
95             printf("NO\n");
96     }
97     return 0;
98 }

#include<cstdio>
#include<queue>
using namespace std;
const int N = 510;
const int INF = 0x7fffffff;
int g[N][N], n, d[N];
int cnt[N];
bool in_q[N];
//SPFA算法
bool SPFA(int s)
{
    queue<int> q; //建立队列
    for( int i = 1 ; i <= n ; i++ )
    {
        cnt[i] = 0; //初始化扩展计数器
        d[i] = INF; //初始化d值
        in_q[i] = false; //初始化in队列标记
    }
    q.push(s); //起点入队列
    cnt[s]++; //扩展计数器自加
    d[s] = 0; //到起点距离置0
    in_q[s] = true; //标记in队列
    while(!q.empty())
    {
        int v = q.front();
        q.pop(); //队头出队列
        in_q[v] = false; //标记出队列
        for( int i = 1 ; i <= n ; i++ )
        {
            if( g[v][i] < INF && d[v]+g[v][i] < d[i] )
            {
                d[i] = d[v]+g[v][i]; //松弛
                if(!in_q[i]) //如果不在队列中
                {
                    in_q[i] = true; //入队列
                    if( ++cnt[i] >= n )
                        return true;
                    q.push(i);
                }
            }
        }
    }
    return false;
}
int main()
{
    int fn, m, w;
    scanf("%d", &fn);
    while(fn--)
    {
        scanf("%d%d%d", &n, &m, &w);
        for( int i = 1 ; i <= n ; i++ )
            for( int j = i+1 ; j <= n ; j++ )
                g[i][j] = g[j][i] = INF; //初始化邻接矩阵
        while(m--)
        {
            int s, e, t;
            scanf("%d%d%d", &s, &e, &t);
            if( t < g[s][e] ) //处理重边
                g[s][e] = g[e][s] = t;
        }
        while(w--)
        {
            int s, e, t;
            scanf("%d%d%d", &s, &e, &t);
            g[s][e] = -t;
        }
        printf("%s\n", SPFA(1)?"YES":"NO"); //只试起点1
    }
    return 0;
}

posted on 2021-01-31 19:24 程帅霞阅读(102) 评论(0) 编辑收藏举报

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