hdu 4948 Kingdom

题目链接

Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 194    Accepted Submission(s): 106
Special Judge

Problem Description

Teacher Mai has a kingdom consisting of n cities. He has planned the transportation of the kingdom. Every pair of cities has exactly a one-way road.

He wants develop this kingdom from one city to one city.

Teacher Mai now is considering developing the city w. And he hopes that for every city u he has developed, there is a one-way road from u to w, or there are two one-way roads from u to v, and from v to w, where city v has been developed before.

He gives you the map of the kingdom. Hope you can give a proper order to develop this kingdom.

 

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains an integer n (1<=n<=500).

The following are n lines, the i-th line contains a string consisting of n characters. If the j-th characters is 1, there is a one-way road from city i to city j.

Cities are labelled from 1.

 

Output

If there is no solution just output "-1". Otherwise output n integers representing the order to develop this kingdom.

 

Sample Input

   
   
   
   

    
    
    
    3
011
001
000
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1 2 3
   
   
   
   
   
   
   
   

    
    
    
    题意：n个城市，一对城市之间有且仅有一条单向边，现在一个城市一个城市的发展，每发展一个城市，必须保证前面发展过的城市到它不超过两条边，也就是说要么与它直接相连，要么与它通过一个点间接相连。让你输出发展城市的先后顺序，无解输出-1.
   
   
   
   
   
   
   
   

    
    
    
    题解：容易想出一种解法，每次都找一个到所有没发展过的点的距离不超过2的点发展，如果不能找到则无解。但是如果暴力求到一个点距离小于等于2的点很可能超时。这题还有更好的做法。首先，因为两点之间一定有一条单向边，所以问题一定是有解的。假设u到v的距离大于2，那么v到u一定有一条边，也就是说v到u的距离一定为1，任意两点之间其中一个一定可以做另一个的前驱，所以问题一定有解。然后，每次找一个点到所有剩下的点的距离不超过2，这个点一定是所有点中出度最大的点。我们可以用反证法证明。假设u为出度最大的点，w1，w2，，wk与u直接相连，假设存在一个点v，使u到v的距离大于2，那么v一定有也与w1，w2，，，wk直接相连，并且v到u一定有条单向边，所以这与假设u为出度最大的点矛盾，所以出度最大的点，到所有点的距离不超过2。代码如下：
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
#include<map>
#define nn 510
#define inff 0x3fffffff
typedef long long LL;
using namespace std;
int n;
char tu[nn][nn];
bool vis[nn];
int a[nn];
int ans[nn];
int main()
{
    int i,j,k;
    while(scanf("%d",&n)&&n)
    {
        for(i=0;i<n;i++)
        {
            scanf("%s",tu[i]);
        }
        memset(a,0,sizeof(a));
        memset(vis,false,sizeof(vis));
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(tu[i][j]=='1')
                    a[i]++;
            }
        }
        int la=0;
        int mx,id;
        for(i=0;i<n;i++)
        {
            mx=-inff;
            for(j=0;j<n;j++)
            {
                if(!vis[j]&&a[j]>mx)
                {
                    mx=a[j];
                    id=j;
                }
            }
            vis[id]=true;
            ans[la++]=id;
            for(k=0;k<n;k++)
            {
                if(tu[k][id]=='1')
                    a[k]--;
            }
        }
        for(i=0;i<la;i++)
        {
            printf("%d%c",ans[i]+1,i==la-1?'\n':' ');
        }
    }
    return 0;
}