浙江财经大学
信息管理与工程学院

HDU 1102 Constructing Roads(Prime)

本文由 Ocrosoft 于 2016-07-30 21:52:38 发表

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20635    Accepted Submission(s): 7879

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
 

Solution

题意:给出一个邻接矩阵表示的村子间的距离,接下来给出m组a b数据,表示ab村子间的路已经造好。

和畅通工程系列中的一题类似,把mp[u][v]和mp[v][u]都标记成0就可以了,而且数据保证矩阵除了i==j都是大于等于1的。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 100+10
using namespace std;
int n,m;
int mp[MAXN][MAXN],lowcast[MAXN];
bool vis[MAXN];
int prime(int u0)
{
    ms(vis),vis[u0]=1;
    int pos=u0,sum=0;
    for(int i=1;i<=n;i++)
        lowcast[i]=(i==pos?0:mp[pos][i]);
    for(int i=1;i<n;i++)
    {
        int minn=INF;
        for(int j=1;j<=n;j++)
            if(lowcast[j]<minn&&!vis[j])pos=j,minn=lowcast[j];
        vis[pos]=1;
        sum+=minn;
        for(int j=1;j<=n;j++)
            if(!vis[j]&&lowcast[j]>mp[pos][j])lowcast[j]=mp[pos][j];
    }
    return sum;
}
int main()
{
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        scanf("%d",&mp[i][j]);
        cin>>m;
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mp[u][v]=mp[v][u]=0;
        }
        printf("%d\n",prime(1));
    }
    return 0;
}

 

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