ural 1982. Electrification Plan -最小生成树

1982. Electrification Plan

Time limit: 0.5 second
Memory limit: 64 MB

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 _ij 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.

Input

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 _ij} (0 ≤ c _ij ≤ 10 ⁵). It is guaranteed that c _ij = c _ji, c _ij > 0 for i ≠ j, c _ii = 0.

Output

Output the minimum cost to electrify all the cities.

Sample

input	output
4 2 1 4 0 2 4 3 2 0 5 2 4 5 0 1 3 2 1 0	3

Problem Author: Mikhail Rubinchik
Problem Source: Open Ural FU Championship 2013

Tags: graph theory   ( hide tags for unsolved problems )

题意：

n个城市，其中k个城市有发电站，问，怎么才能让每个城市都有电，并且花费最小

解题思路

每个有发电站的城市都能做出一个最小生成树来，然后好几个最小生成树。但是这样不好写

我们设置一个超级发电站，一开始所有发电站都与之相连，让超级发电站当这课树的根，

这样，既达到了将好几棵树变成一棵树的目的，又使得发电站与发电站之间边的权值不会被加进结果去

然后用 kruskl 算法做最小生成树，

这里注意，因为是从边权值最小开始，权值最小的边的节点的父亲可能都不是超级发电站，

所以后期遇到一个以超级发电站为父亲节点的，要让不是超级发电站为父亲节点的连进去。（有点乱，见代码）

代码

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 110;
int n,k;
int fa[maxn];
int suroot;
int root[maxn];
void init(){
    for(int i = 0;i