SDAU 练习四1001 acm村庄建设最小长度问题

                                                   Problem A

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

[align=left]Input[/align]
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.<br><br>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.<br>
 

[align=left]Output[/align]
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. <br>
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

179

简单题意：

                          有N个村庄,从1到N编号,你应该建立一些道路,这样每一个两个村庄可以连接到对方。我们说两个村庄A和B是连接,当且仅当有A和B之间的道路,或存在C这样的一个村庄之间有一条路和C,C和B连接。我们知道已经有一些道路之间的一些村庄,你的工作是构建道路,所有的村庄都连接和道路建设是最小的长度。

简单思路：

        采用邻接表的方式将数据保存下来。在用邻接表优化的prim算法即可。。典型的例题哈  

4000
ACID:00776486

代码如下：

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;

const int maxn = 101;

struct Edge

{

    int begin;

    int end;

    int weight;//边的权值

}edges[maxn*maxn];

int father[maxn];

int map[maxn][maxn];

int find(int a)

{

    if(a == father[a])

        return a;

    return father[a] = find(father[a]);

}

int kruscal(int count)

{

    int i;

    for(i = 1 ; i < maxn ; ++i)

        {

            father[i] = i;

        }

    int sum = 0;

    for(i = 1 ; i <= count ; ++i)

        {

           int fx = find(edges[i].begin);

           int fy = find(edges[i].end);

           if(fx != fy)

            {

                father[fx] = fy;

                sum += edges[i].weight;

            }

        }

    return sum;

}

bool cmp(const Edge& a, const Edge& b)

{

    return a.weight < b.weight;

}

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        int i;

        int j;

        for(i = 1 ; i <= n ; ++i)

        {

            for(j = 1 ; j <= n ; ++j)

            {

                cin>>map[i][j];

            }

        }

        int m;

        cin>>m;

        while(m--)

        {

            int a,b;

            cin>>a>>b;;

            map[a][b] = map[b][a] = 0;

        }

        int cnt = 1;

        for(i = 1 ; i <= n ; ++i)

            {

                for(j = 1 ; j <= i ; ++j)

                {

                    edges[cnt].begin = i;

                    edges[cnt].end = j;

                    edges[cnt++].weight = map[i][j];

                }

            }

        cnt = cnt-1;

        sort(edges+1,edges+1+cnt,cmp);

        cout<<kruscal(cnt)<<endl;

    }

    return 0;

}