（最短路）SPOJ - Ada and Cycle （易）-优快云博客

本文介绍了一个有趣的图论问题——在Bugindia的地图上寻找从每个城市出发并回到该城市的最短路径长度。通过使用基础的广度优先搜索（BFS）算法，我们能够解决这个问题，并提供了一段实现这一算法的C++代码示例。

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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 ≤ H_ij ≤ 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

Example Output

2
2
1
2
NO WAY

Example Input

Example Output

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 }