Timusoj 1982. Electrification Plan

http://acm.timus.ru/problem.aspx?space=1&num=1982

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 incij 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 × ntable of integers {cij} (0 ≤ cij ≤ 105). It is guaranteed that cij = cji, cij > 0 for i ≠ j, cii = 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  (

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Difficulty: 144    Printable version    Submit solution    Discussion (4)
All submissions (2430)    All accepted submissions (894)    Solutions rating (630)

分析：

无向图，给n个点，n^2条边，每条边有个一权值，其中有k个点有发电站，给出这k个点的编号，选择最小权值的边，求使得剩下的点都能接收到电。

　　发电站之间显然不能有边，那么把k个点合成一个点，然后在图上就MST就可以了。

AC代码1：

　　　　1、edge[i][j]=0是个很巧妙的设置。

            2、求最小生成树，由于生成树是图的极小联通子图。最小生成树一定要包含图中所有的点。

#include<cstdio>
#include<iostream>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;

int edge[110][110];
int vis[110],dis[110];
bool flag[110];
int n,k,ans;

void prim()
{
    int u=1,minw;
    for(int i=1;i<=n;i++)
    {
        vis[i]=0;
        dis[i]=edge[u][i];
    }
    vis[u]=1;
    for(int i=1;i<n;i++)
    {
        minw=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && dis[j]<minw)
            {
                minw=dis[j];
                u=j;
            }
        }
        ans+=minw;
        vis[u]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && edge[u][j]<dis[j])
                dis[j]=edge[u][j];
        }
    }
}

int main()
{
    int d;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        memset(flag,false,sizeof(flag));
        ans=0;
        for(int i=0;i<k;i++)
        {
            scanf("%d",&d);
            flag[d]=true;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&edge[i][j]);
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(flag[i]&&flag[j])
                    edge[i][j]=0;
            }
        }
        prim();
        printf("%d
",ans);
    }
    return 0;
}

AC代码2：

//STATUS:C++_AC_31MS_401KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=110;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v,val;
    bool operator < (const Edge& a)const {
        return val<a.val;
    }
}e[N*N];
int n,k;
int id[N],p[N],w[N][N];

int find(int x){return p[x]==x?x:p[x]=find(p[x]);}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,a,x,y,ans,cnt;
    while(~scanf("%d%d",&n,&k))
    {
        mem(id,0);
        for(i=0;i<k;i++){
            scanf("%d",&a);
            id[a]=1;
        }
        k=2;
        for(i=1;i<=n;i++){
            if(id[i])continue;
            id[i]=k++;
        }
        mem(w,INF);
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&a);
                w[id[i]][id[j]]=Min(w[id[i]][id[j]],a);
            }
        }
        cnt=0;
        for(i=1;i<k;i++){
            for(j=i+1;j<k;j++){
                e[cnt].u=i,e[cnt].v=j;
                e[cnt].val=w[i][j];
                cnt++;
            }
        }
        sort(e,e+cnt);
        ans=0;
        for(i=1;i<k;i++)p[i]=i;
        for(i=0;i<cnt;i++){
            x=find(e[i].u);y=find(e[i].v);
            if(x!=y){
                p[y]=x;
                ans+=e[i].val;
            }
        }

        printf("%d
",ans);
    }
    return 0;
}