最小生成树

输入

输入的第一行包含一个正整数n，表示图中共有n个顶点。其中n不超过50。

以后的n行中每行有n个用空格隔开的整数，对于第i行的第j个整数，如果不为0，则表示第i个顶点和第j个顶点有直接连接且代价为相应的值，0表示没有直接连接。当i和j相等的时候，保证对应的整数为0。

输入保证邻接矩阵为对称矩阵，即输入的图一定是无向图，且保证图中只有一个连通分量。

输出

只有一个整数，即最小生成树的总代价。请注意行尾输出换行。

样例输入

4 0 2 4 0 2 0 3 5 4 3 0 1 0 5 1 0

样例输出

6

 1 #include <stdio.h>
#define N 55
#define INF 600000
#define INF2 600001
typedef struct g{
    int arc[N][N];
    int arcNum,verNum;
    int maxWeight;
}Graph;
typedef struct {
    int adjvex;
    int lowcost;
}Closedge;
void Create(Graph *G,int n){
    int i,j;
    G->verNum = G->arcNum = n;
    G->maxWeight = 0;
    for(i = 0; i < G->arcNum; i++)
        for(j = 0; j < G->arcNum; j++){
            scanf("%d",&G->arc[i][j]);
            if(G->maxWeight < G->arc[i][j])
                G->maxWeight = G->arc[i][j];
        }
}
void Prim(Graph *G,int &spent){
    Closedge closedge[N];
    int i,j,k,min,minVer;
    //初始化数组closedge
    for(i = 0; i < G->verNum; i++){
        closedge[i].adjvex = 0;//从结点 0 开始
        if(G->arc[0][i])
            closedge[i].lowcost = G->arc[0][i];
        else
            closedge[i].lowcost = INF;
    }
 
    closedge[0].lowcost = INF2;
     
    for(j = 0; j < G->verNum - 1; j++){
        min = G->maxWeight;
        //选择最小的边
        for(k = 0; k < G->verNum; k++)
            if(closedge[k].lowcost && min > closedge[k].lowcost){
                min = closedge[k].lowcost;
                minVer = k;//closedge[k].adjvex
            }
        spent += min;
        closedge[minVer].lowcost = INF2;//加入入选集合
        //查找顶点minVer到没入选顶点的“花费” minVer = 1;
        for(k = 0; k < G->verNum; k++){
            if(closedge[k].lowcost != INF2 && G->arc[minVer][k] < closedge[k].lowcost && G->arc[minVer][k]){
                closedge[k].adjvex = k;
                closedge[k].lowcost = G->arc[minVer][k];
            }
        }
    }
}
int main(){
    int n;
    Graph G;
    scanf("%d",&n);
    Create(&G,n);
    int m = 0;
    Prim(&G,m);
    printf("%d
",m);
    return 0;
}