# POJ 1861 Network C语言版 朴素Prim

POJ题解 1543浏览

Prim算法的应用

Description

Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.

Input

The first line of the input contains two integer numbers: N – the number of hubs in the network (2 <= N <= 1000) and M – the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections – the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.

Output

Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P – the number of cables used, then output P pairs of integer numbers – numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.

Sample Input

4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1

Sample Output

1
4
1 2
1 3
2 3
3 4

Slyar:WA了好几次才过，以后再也不想做Special Judge的题了，晕死…说下题意，给出节点个数m和边数n，下面n行给出边(x,y)以及权值w。输出第一行为最小生成树中的最大边权值，第二行为 一个可行生成树方案的边数k，下面k行为可行生成树的k条边。题目是Special Judge，意思就是不具有唯一解，可能有多解，样例输出为以下结果也可算对。

1
3
1 3
2 4
2 3

```#include
#include

#define MAX 10005
#define MAXCOST 0x7fffffff

int graph[MAX][MAX];
int max = 0;

struct
{
int x, y;
}p[MAX];

int Prim(int graph[][MAX], int n)
{

int lowcost[MAX];
int mst[MAX];
int i, j, min, minid, sum = 0;
for (i = 2; i <= n; i++)
{
lowcost[i] = graph[i];
mst[i] = 1;
}
mst = 0;
for (i = 2; i <= n; i++)
{
min = MAXCOST;
minid = 0;
for (j = 2; j <= n; j++)
{
if (lowcost[j] < min && lowcost[j] != 0)
{
min = lowcost[j];
minid = j;
}
}
sum++;
p[sum].x = mst[minid];
p[sum].y = minid;
if (lowcost[minid] > max)
{
max = lowcost[minid];
}
lowcost[minid] = 0;
for (j = 2; j <= n; j++)
{
if (graph[minid][j] < lowcost[j])
{
lowcost[j] = graph[minid][j];
mst[j] = minid;
}
}
}
return sum;
}

int main()
{
int i, j, k, m, n;
int cost;
scanf("%d%d", &m, &n);
for (i = 1; i <= m; i++)
{
for (j = 1; j <= m; j++)
{
graph[i][j] = MAXCOST;
}
}
for (k = 0; k < n; k++)
{
scanf("%d %d %d", &i, &j, &cost);
graph[i][j] = cost;
graph[j][i] = cost;
}
n = Prim(graph, m);
printf("%d\n", max);
printf("%d\n", n);
for (i = 1; i <= n; i++)
{
printf("%d %d\n", p[i].x, p[i].y);
}
//system("pause");
return 0;
}
```