浙江财经大学
信工学院ACM集训队

HDU 1102 Constructing Roads(Kruskal)

本文由 Ocrosoft 于 2016-07-30 23:17:04 发表

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20647    Accepted Submission(s): 7881

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

Prime算法传送门:不传送了,上一篇就是。

题意:传送大法↑

edge数组开小了害我整整TLE了一晚上…

#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;
struct Edge
{
    int u,v,w;
    bool operator <(const Edge b) const
    {
        return w<b.w;
    }
}edge[10010];
int pre[MAXN];
int find(int x)
{
    if(x!=pre[x])pre[x]=find(pre[x]);
    return pre[x];
}
void Union(int x,int y)
{
    int fx=find(x),fy=find(y);
    if(fx!=fy)pre[fx]=fy;
}
int main()
{
    while(cin>>n)
    {
        int cur=1;
        for(int i=1;i<=n;i++)pre[i]=i;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        if(i!=j)edge[cur].u=i,edge[cur].v=j,scanf("%d",&edge[cur].w),cur++;
        else scanf("%*d");
        cin>>m;
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            Union(u,v);
        }
        m=cur-1;
        sort(edge+1,edge+1+m);
        int sum=0;
        for(int i=1;i<=m;i++)
        {
            int fx=find(edge[i].u),fy=find(edge[i].v);
            if(fx!=fy)pre[fx]=fy,sum+=edge[i].w;
        }
        printf("%d\n",sum);
    }
    return 0;
}

 

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