练习四 Problem1001

题目：

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. <br><br>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.<br>

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.<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>

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. <br>

Sample Input

30 990 692990 0 179692 179 011 2

Sample Output

179

题意：有N个村庄，你需要建造道路使所有的村庄可以连在一起，如果有ABC三个村庄，AB之间有一条道路，AC之间也有一条道路，那么BC之间因为这两条线也能连在一起。我们已经知道有的村庄之间有已经建造的道路，求使所有村庄连在一起需要再建造的道路的最小值。

解题思路：这个题是一个关于最小生成树的问题，题目给定一些已经建造好的道路。每条边有三个属性，起点、终点和权值，如果起点和终点相反，只记录一个就可以，然后用一个数组将每个顶点初始化为1到n，如果已经建造好的道路则把起点和终点都置成起点值，然后调用自己写的Prim函数，求出建造的最小值。

感悟：刚接触数据结构，一切都是那么陌生。最小生成树学了两种方法，prim算法和kruskal算法。在纸上找到一个图的最小生成树不难，但在写代码的时候却是十分的不容易。

代码：

<span style="font-size:14px;">#include<stdio.h>#define MAX 1000000int map[105][105];int lowcost[105];int N;void Prim(int map[][105],int n){    int i,j,k,road=0;    int min;    for(i=2;i<=N;i++)            lowcost[i]=map[1][i];        for(i=2;i<=N;i++)        {            min=lowcost[i];            k=i;            for(j=2;j<=N;j++)                if(min>lowcost[j])                {                    min=lowcost[j];                    k=j;                }            road+=min;            lowcost[k]=MAX;            for(j=2;j<=N;j++)            {                if((map[k][j]<lowcost[j])&&(lowcost[j]<MAX))                {                    lowcost[j]=map[k][j];                }            }        }        printf("%d\n",road);}int main(){    int n,i,j;    while(scanf("%d",&N)!=EOF)    {                for(i=1;i<=N;i++)            for(j=1;j<=N;j++)                scanf("%d",&map[i][j]);        scanf("%d",&n);        while(n--)        {            scanf("%d%d",&i,&j);            map[i][j]=map[j][i]=0;                    }        Prim(map,N);       }    return 0;}</span>