Ada and Cycle

最新推荐文章于 2025-06-09 15:21:33 发布

原创最新推荐文章于 2025-06-09 15:21:33 发布 · 366 阅读

0 ·

CC 4.0 BY-SA版权

最短路专栏收录该内容

1 篇文章

订阅专栏

在Bugindia的多个城市间，利用邻接矩阵表示城市间的单向道路，通过基础BFS算法寻找从每个城市出发并返回的最短路径长度。如果不存在这样的路径，则输出特定提示。

ADACYCLE - Ada and Cycle

no tags

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

Example Output

2
2
1
2
NO WAY

Example Input

Example Output

Submit solution!

2.题意概述：

给你邻接矩阵，其中ij元素的0表示i到j没有道路，1表示有长度为1的道路，某人喜欢短途旅行，要你判断他在i城市时候的最短旅行路程是什么，不存在则输出NO WAY

3.解题思路

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。

4.AC代码：

#include<bits/stdc++.h>
using namespace std;
const int N=2222;
const int inf=0x3f3f3f3f;
vector<int>vv[N];
bool vis[N];
int ans;
int n;
typedef pair<int,int> p;


int bfs(int i)
{
    queue<p>q;
    q.push(make_pair(0,i));

    while(!q.empty())
    {
        int top=q.front().second;
        int dis=q.front().first;
        q.pop();
        for(int j=0;j<vv[top].size();j++)
        {
            int tem=vv[top][j];
            if(!vis[tem])
            {
                vis[tem]=1;
                if(tem==i)
                    return  dis+1;
                    q.push(make_pair(dis+1,tem));
            }

        }
    }
    return inf;
}
int main()
{
    cin>>n;
    int tem;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&tem);
            if(tem)
            {
                vv[i].push_back(j);
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        ans=inf;
        if(vv[i].size())
        {
            memset(vis,false,sizeof(vis));
            ans=bfs(i);
        }
        if(ans==inf)
        {
            puts("NO WAY");
        }
        else
        {
            printf("%d\n",ans);
        }
    }
}