<poj-2485>Highways

比眉伴天荒 2022-07-14 04:11 57阅读 0赞

Highways

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;29427</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;13402</td> 
  </tr> 
 </tbody> 
</table>

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible to drive between any pair of towns without leaving the highway system.  
  
Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.  
  
The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.

Input

The first line of input is an integer T, which tells how many test cases followed.  
The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within \[1, 65536\]) between village i and village j. There is an empty line after each test case.

Output

For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.

Sample Input

1
    
    3
    0 990 692
    990 0 179
    692 179 0

Sample Output

692

解题思路

其实我真的没有真正的看懂这个题

现在觉得读题越来越吃力，还是得好好学英语

就是求最小生成树的最大权值

#include<stdio.h>
    #include<string.h>
    #define MAX 0x3f3f3f3f
    
    int mp[500][500];
    int dis[500];
    int vis[500];
    
    int prim(int n)
    {
        int mi, minin;
        int k = 0;
        dis[1] = 0;
        for(int i = 1; i <= n; i++)
        {
            mi = MAX;
            for(int j = 1; j <= n; j++)//找权值最小的边
            {
                if(vis[j]==0&&dis[j] != -1&&dis[j] < mi)
                {
                    mi = dis[j];
                    minin = j;
                }
            }
            if(mi > k)//与现在权值最大的比较
            {
                k = mi;
            }
            vis[minin] = 1;
            for(int j = 1; j <= n; j++)//修改与已加入节点的最小权值
            {
                if(vis[j]==0&&(dis[j] == -1||dis[j] > mp[minin][j]))
                {
                    dis[j] = mp[minin][j];
                }
            }
        }
        return k;
    }
    
    int main()
    {
        int t, n;
        scanf("%d",&t);
        while(t--)
        {
            memset(mp, 0, sizeof(mp));
            memset(vis, 0, sizeof(vis));
            memset(dis, -1, sizeof(dis));
            scanf("%d", &n);
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= n; j++)
                {
                    scanf("%d",&mp[i][j]);
                }
            }
            printf("%d\n",prim(n));
        }
        return 0;
    }