POJ 2485:Highways(最小生成树&&prim)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21628 | Accepted: 9970 |
Description
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 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
Sample Input
1 3
0 990 692
990 0 179
692 179 0
Sample Output
692
这题写的真烦。
。
各种不知道的莫名其妙的
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath> using namespace std; const int INF=1000000000;
const int N=510;
int map[N][N];
int vis[N];
int ans;
int dis[N]; void prim(int n)//prim求最小生成树
{
memset(vis, 0, sizeof(vis));
for(int i=1; i<=n; i++)
{
dis[i] = INF;
}
dis[1] = 0;
ans = 0;
for(int i=1; i<=n; i++)
{
int temp = INF, k = 0;
for(int j=1; j<=n; j++)
{
if(!vis[j] && dis[j]<temp)
{
temp = dis[j];
k = j;
}
}
vis[k] = true;
if(ans<temp) ans = temp;
for(int j=1; j<=n; j++)
{
if(!vis[j] && dis[j]>map[k][j])
{
dis[j] = map[k][j];
}
}
}
} int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int n;
scanf("%d", &n);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
scanf("%d", &map[i][j]);
}
}
prim(n);
printf("%d\n", ans);
}
return 0;
}
错误。。
版权声明:本文博客原创文章,博客,未经同意,不得转载。
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