Highways
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 14220 |
|
Accepted: 6620 |
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
Hint
Huge input,scanf is recommended.
這次還是1A，爽歪歪啊，也沒想到
那個這個題求的是讓任意兩個城市之間有路徑
并且使得最長的公路的距離最短
所以應(yīng)該讓任意兩個城市之間的路盡量短，這符合最小生成樹的性質(zhì)
還是用的該過的prim的代碼，稍微修改了下，還是矩陣存的，費內(nèi)存啊
其實我覺得記錄邊的話，并查集優(yōu)化的kruskal可能更好用
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 600
5int map[MAX][MAX],ans;
6int bian[1000][2],v[MAX];
7int n;
8short vis[MAX];
9void init()
10
{
11
int i,j;
12 scanf("%d",&n);
13 for (i=1;i<=n ;i++ )
14
{
15 for (j=1;j<=n ;j++ )
16 {
17 scanf("%d",&map[i][j]);
18
}
19 }
20
}
21void prim()
22{
23 int i,j,k,dmin;
24 memset(vis,0,sizeof(vis));
25 vis[1]=1;
26 v[1]=1;
27 for (i=2;i<=n;i++)
28 {
29 dmin=0x7fffffff;
30 for (k=1;k<=i-1 ;k++ )
31 {
32 for (j=1;j<=n ;j++ )
33 if ((!vis[j])&&(j!=v[k])&&(map[v[k]][j]<dmin))
34 {
35 dmin=map[v[k]][j];
36 v[i]=j;
37 bian[i-1][0]=v[k];
38 bian[i-1][1]=j;
39 }
40 }
41 vis[v[i]]=1;
42 }
43 ans=0;
44 for (i=1;i<=n-1 ;i++ )
45 if (ans<map[bian[i][0]][bian[i][1]])
46 {
47 ans=map[bian[i][0]][bian[i][1]];
48 }
49}
50int main()
51{
52 int t;
53 scanf("%d",&t);
54 while (t)
55 {
56 init();
57 prim();
58 printf("%d\n",ans);
59 t--;
60 }
61 return 0;
62}
63