HDU 1102 Constructing Roads

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5227    Accepted Submission(s): 1896

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected. 

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

 

 

Input

The first line is an integer N (3 <= N <= 100), 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, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

 

 

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum. 

 

 

Sample Input

3 0 990 692 990 0 179 692 179 0 1 1 2

 

 

Sample Output

179

 

很简单的最小生成树，有两种做法，我采用Kruskal和Prim都试了一下。一种是使已经有的边距离为0，这样既能顺利地生成，又能不影响答案；第二种是合并已经有的两条边的集合。我这里kruskal采用的是第二种方法，Prim只能用第一种方法。

Kruskal Algorithm Code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 107

int mp[N][N],fa[N],vis[N][N],n,res;

struct Edge
{
    int s,t,w;
}edge[N*N];

int cmp(Edge ka,Edge kb)
{
    return ka.w < kb.w;
}

int findset(int x)
{
    if(x != fa[x])
        fa[x] = findset(fa[x]);
    return fa[x];
}

void Kruskal()
{
    int i,j,k = 0,q,A,B;
    for(i=1;i<=n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            edge[k].s = i;
            edge[k].t = j;
            edge[k].w = mp[i][j];
            k++;
        }
    }
    sort(edge,edge+k,cmp);
    for(i=1;i<=n;i++)
        fa[i] = i;
    res = 0;
    scanf("%d",&q);
    for(i=0;i<q;i++)
    {
        scanf("%d%d",&A,&B);
        int u = findset(A);
        int v = findset(B);
        fa[u] = v;
    }
    for(i=0;i<k;i++)
    {
        int u = edge[i].s;
        int v = edge[i].t;
        int fx = findset(u);
        int fy = findset(v);
        if(fx != fy)
        {
            res += edge[i].w;
            fa[fx] =fy;
        }
    }
}

int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&mp[i][j]);
        Kruskal();
        printf("%d\n",res);
    }
    return 0;
}

 

Prim Algorithm Code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define Mod 1000000007
using namespace std;
#define N 107

int mp[N][N],vis[N],n,res,len[N];

void Prim()
{
    int i,j,k,mini;
    res = 0;
    memset(vis,0,sizeof(vis));
    for(i=1;i<=n;i++)
        len[i] = mp[1][i];
    len[1] = 0;
    vis[1] = 1;
    for(i=1;i<=n;i++)
    {
        mini = Mod;
        for(j=1;j<=n;j++)
        {
            if(!vis[j] && len[j] < mini)
            {
                mini = len[j];
                k = j;
            }
        }
        if(mini == Mod)
            return;
        res += len[k];
        vis[k] = 1;
        for(j=1;j<=n;j++)
        {
            if(!vis[j] && len[j] > mp[k][j])
                len[j] = mp[k][j];
        }
    }
}
int main()
{
    int i,j,q,u,v;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&mp[i][j]);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d",&u,&v);
            mp[u][v] = mp[v][u] = 0;
        }
        Prim();
        printf("%d\n",res);
    }
    return 0;
}