zufeoj Electrification Plan （最小生成树，巧妙设e[i][j]=0）

Electrification Plan

时间限制: 1 Sec  内存限制: 128 MB
提交: 31  解决: 13
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题目描述

Some country has n cities. The government has decided to electrify all these cities. At first, power stations in k different cities were built. The other cities should be connected with the power stations via power lines. For any cities i, j it is possible to build a power line between them in c[i][j] roubles. The country is in crisis after a civil war, so the government decided to build only a few power lines. Of course from every city there must be a path along the lines to some city with a power station. Find the minimum possible cost to build all necessary power lines.

输入

多组测试数据。

The first line contains integers n and k (1 ≤ k ≤ n ≤ 100). The second line contains k different integers that are the numbers of the cities with power stations. The next n lines contain an n × n table of integers c[i][j] (0 ≤ c[i][j] ≤ 10^5). It is guaranteed that c[i][j] = c[j][i], c[i][j] > 0 for i ≠ j, c[i][i] = 0.

输出

Output the minimum cost to electrify all the cities.

样例输入

4 2
1 4
0 2 4 3
2 0 5 2
4 5 0 1
3 2 1 0

样例输出

3

输入的时候很巧妙，要把发电站相互之间的权值设为0

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#define inf 0x3f3f3f3f;
using namespace std;
int n, k;
int di[];
int o;
int pra[];
int sum = ;
struct node
{
    int u, v, w;
}a[];
bool cmp(node a, node b)
{
    return a.w < b.w;
}
int find(int x)
{
    if (pra[x] == x) return x;
    else return pra[x] = find(pra[x]);
}
bool same(int x,int y)
{
    return find(x) == find(y);
}
void unite(int xx,int yy)
{
    int x = find(xx);
    int y = find(yy);
    if (x == y) return;
    else pra[x] = y;
}
void kruskal()//套模版
{
    sum = ;
    int i;
    sort(a + , a + o, cmp);
    for (i = ; i <= o - ;i++)
    {
        node x = a[i];
        if (same(a[i].u, a[i].v)) continue;
        else
        {
            sum = sum + a[i].w;
            unite(a[i].u, a[i].v);
        }
    }
}
int main()
{
    while (cin >> n >> k)
    {
        int i, j;
        for (i = ; i <= k; i++)
        {
            cin >> di[i];//发电站
        }
        o = ;
        for (i = ; i <= k; i++)
        {
            for (j = ; j <= k; j++)
            {//把发电站之间的w设为0就可以了
                a[o].u = di[i];
                a[o].v = di[j];
                a[o++].w = ;
            }
        }
        for (i = ; i <= n; i++)
        {
            pra[i] = i;
            for (j = ; j <= n; j++)
            {
                a[o].u = i;
                a[o].v = j;
                cin >> a[o++].w;
            }
        }
        kruskal();
        cout << sum << endl;
    }
}