首页 > ACM题库 > HDU-杭电 > HDU 1102 Constructing Roads-最小生成树-[解题报告] C++
2013
11-27

HDU 1102 Constructing Roads-最小生成树-[解题报告] C++

Constructing Roads

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

179

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1102

 

一道入门的最小生成树题目。

这道题目描述的边是双向的,xh[i][j]=xh[j][i],所以我们要把它变成单向的,读数据加 if(i<j) 就增加一条边,而下面Q条已经修好的路也要判断(edge[i][j],始终保持i<j)。

这道题数据量不是很大,所以find函数里面的剪枝可有可无。

 

代码如下:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
using namespace std;

/*
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
*/

int n,m;//n个点,总共有m跳边,单向边
int xh[105][105];
int parent[105];//父亲节点
struct se
{
    int x,y,w;
}edge[5055];

bool emp(se a,se b)
{
    return a.w<b.w;
}

void UFset()
{
    for(int i=0;i<=n;i++)
        parent[i]=i;
}

int find(int x)
{
    int r=x;
    while(x!=parent[x])
        x=parent[x];
    while(r!=x)//剪枝
    {
        int j=parent[r];
        parent[r]=x;
        r=j;
    }
    return x;
}

void Kruskal()
{
    int sum=0,i,count=0;
    UFset();
    for(i=0;i<m;i++)
    {
        int x=edge[i].x;
        int y=edge[i].y;
        int fx=find(x),fy=find(y);
        if(fx!=fy)
        {
            count++;
            parent[fx]=fy;
            sum+=edge[i].w;
            if(count==n-1)
                break;
        }
    }
    cout<<sum<<endl;
}

int main()
{
    int i,j,k,x;
    while(cin>>n)
    {
        m=0;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&x);
                if(i<j)//因为是单向的,所以只用i<j右上角那一边就行了
                {
                    edge[m].x=i;
                    edge[m].y=j;
                    edge[m].w=x;
                    xh[i][j]=m;//记录i和j是第几条边
                    m++;
                }
            }
        cin>>k;
        while(k--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            if(a>b)//和上面一样,始终保持edge[i][j],i<j
                swap(a,b);
            edge[xh[a][b]].w=0;//已经有路了,w就记为0
        }
        sort(edge,edge+m,emp);//边从小到大排序
        Kruskal();
    }
    return 520;
}

 


  1. I like your publish. It is great to see you verbalize from the coronary heart and clarity on this essential subject matter can be easily noticed.

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  3. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。

  4. /*
    * =====================================================================================
    *
    * Filename: 1366.cc
    *
    * Description:
    *
    * Version: 1.0
    * Created: 2014年01月06日 14时52分14秒
    * Revision: none
    * Compiler: gcc
    *
    * Author: Wenxian Ni (Hello World~), [email protected]
    * Organization: AMS/ICT
    *
    * =====================================================================================
    */

    #include
    #include

    using namespace std;

    int main()
    {
    stack st;
    int n,i,j;
    int test;
    int a[100001];
    int b[100001];
    while(cin>>n)
    {
    for(i=1;i>a[i];
    for(i=1;i>b[i];
    //st.clear();
    while(!st.empty())
    st.pop();
    i = 1;
    j = 1;

    while(in)
    break;
    }
    while(!st.empty()&&st.top()==b[j])
    {
    st.pop();
    j++;
    }
    }
    if(st.empty())
    cout<<"YES"<<endl;
    else
    cout<<"NO"<<endl;
    }
    return 0;
    }