（最短路）SPOJ - Ada and Cycle （易）

Ada the Ladybug is on a trip in Bugindia. There are many cities and some uni-directional roads connecting them. Ada is wondering about the shortest path, which begins in a city and ends in the same city. Since Ada likes short trips, she asked you to find the length of such path for each city in Bugindia.

Input

The first line will contain 0 < N ≤ 2000, the number of cities.

Then N lines follow, each containing N integers 0 ≤ Hij ≤ 1. One means, that there is a road between i and j (zero means there isn't a road).

Output

Print N lines, the length of shortest path which begins in city i and ends in city i. If the path doesn't exist, print "NO WAY" instead.

Example Input

5
0 1 1 1 1
1 0 0 0 1
0 0 1 1 0
0 0 1 0 0
0 0 0 1 0

Example Output

2
2
1
2
NO WAY

Example Input

5
0 1 0 0 1
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 0 0 0 0

Example Output

2
5
5
5
2

时间很宽松，因为每条边权值相同都为1，用最基础的bfs即可。

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <stack>
12 #define mp make_pair
13 //#define INF 0x7fffffff
14 typedef long long ll;
15 typedef unsigned long long ull;
16 const int MAX=2e3+5;
17 const int INF=0x7fffffff;
18 using namespace std;
19 typedef pair<int,int> pii;
20 int n;
21 vector <int> g[MAX];
22 queue <pii> que;
23 bool vi[MAX];
24 int tem,an;
25 int bfs(int i)
26 {
27     while(!que.empty())
28         que.pop();
29     que.push(mp(0,i));
30     while(!que.empty())
31     {
32         pii y=que.front();
33         que.pop();
34         int dis=y.first,id=y.second;
35         for(int j=0;j<g[id].size();j++)
36         {
37             int to=g[id][j];
38             if(!vi[to])
39             {
40                 vi[to]=true;
41                 if(to==i)
42                     return dis+1;
43                 que.push(mp(dis+1,to));
44             }
45         }
46     }
47     return INF;
48 }
49 int main()
50 {
51     scanf("%d",&n);
52     for(int i=1;i<=n;i++)
53     {
54         for(int j=1;j<=n;j++)
55         {
56             scanf("%d",&tem);
57             if(tem)
58                 g[i].push_back(j);
59         }
60     }
61     for(int i=1;i<=n;i++)
62     {
63         an=INF;
64         if(g[i].size())
65         {
66             memset(vi,false,n+1);
67             an=bfs(i);
68         }
69         if(an==INF)
70             printf("NO WAY\n");
71         else
72             printf("%d\n",an);
73     }
74 }

 

posted @ 2017-04-07 00:04  perplex  阅读(359)  评论(0)  编辑  收藏  举报