# POJ 2485 Highways C语言版 Kruskal

POJ题解 2819浏览

Kruskal算法的应用

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

Slyar:题意很简单，就是给你t组数据，每组数据给你一个n，然后下面是n*n的无向图邻接矩阵表示，求最小生成树的最大边权。

```#include
#include

#define MAX 500

/* 定义边(x,y)，权为w */
typedef struct
{
int x, y;
int w;
}edge;

edge e[MAX * MAX / 2];
/* father[x]表示x的父节点 */
int father[MAX];
int max;

/* 比较函数，按权值非降序排序 */
int cmp(const void *a, const void *b)
{
return (*(edge *)a).w - (*(edge *)b).w;
}

/* 初始化集合 */
void Make_Set(int x)
{
father[x] = x;
}

/* 查找x元素所在的集合,回溯时压缩路径 */
int Find_Set(int x)
{
if (x != father[x])
{
father[x] = Find_Set(father[x]);
}
return father[x];
}

/* 合并x,y所在的集合 */
void Union(int x, int y, int w)
{

if (x == y) return;
father[y] = x;
if (w > max) max = w;
}

int main()
{
int i, j, n, k, t;
int tmp;

scanf("%d", &t);

while (t--)
{
k = 0;
max = 0;

scanf("%d", &n);

for (i = 0; i < n; i++)
{
Make_Set(i);
for (j = 0; j < n; j++)
{
/* 只读取上三角 */
if (i < j)
{
e[k].x = i;
e[k].y = j;
scanf("%d", &e[k].w);
k++;
}
else
{
scanf("%d", &tmp);
}
}
}

qsort(e, k, sizeof(edge), cmp);

for (i = 0; i < k; i++)
{
Union(Find_Set(e[i].x), Find_Set(e[i].y), e[i].w);
}

printf("%d\n", max);
}
//system("pause");
return 0;
}
```