典型的prim算法
这类题目可以稍作变形,比如POJ2421
#include <iostream>
#include <map>
#define MAXN 102
typedef long elem_t;
using namespace std;
elem_t prim(int n,elem_t mat[MAXN][MAXN]){
elem_t closeEdge[MAXN],sum=0,min;
int i,j,k;
for(i = 0; i < n; i++)
closeEdge[i] = mat[0][i];//初始化辅助数组
for(i = 1; i < n; i++)
{
j = 0;
while(!closeEdge[j])j++;//寻找第一个未归入的点
min = closeEdge[j];k = j;
//开始寻找下一节点
for(j++; j < n; j++)
if(closeEdge[j] && min > closeEdge[j])
{
min = closeEdge[j];
k = j;
}
sum += closeEdge[k];
closeEdge[k] = 0;
for(j = 0; j < n; j++)//新顶点入集并重新选择最小边
if(mat[k][j] && mat[k][j] < closeEdge[j])
closeEdge[j] = mat[k][j];
}
return sum;
}
int main(){
int n,i,j;
long result;
long distance[MAXN][MAXN];
while(cin>>n){
for (i = 0;i < n;i++)
for (j = 0;j < n;j++){
cin>>distance[i][j];
}
result = prim(n,distance);
cout<<result<<endl;
}
return 0;
}