依旧是最小生成树的基础题~
#include<iostream>
#include<algorithm>
using namespace std;
int father[101];
struct n1
{
int s,e,w;
};
n1 path[6000];
bool cmp(n1 a,n1 b)
{
return a.w<b.w;
}
int find(int i)
{
while(father[i]!=i)
{
i=father[i];
}
return i;
}
int kruskal(int v,int e)
{
int temp1,temp2,i,sum,num_bian;
for(i=1;i<=v;i++)
{
father[i]=i;
}
sort(path+1,path+e+1,cmp);
i=sum=num_bian=0;
v--;
while(num_bian!=v)
{
i++;
temp1=find(path[i].s);
temp2=find(path[i].e);
if(temp1!=temp2)
{
sum+=path[i].w;
num_bian++;
if(temp1>temp2)
{
temp1^=temp2^=temp1^=temp2;
}
father[temp2]=temp1;
}
}
return sum;
}
int main()
{
int t,i,j,k,map[101][101],n;
while(scanf("%d",&n)!=EOF){
k=0;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
scanf("%d",&map[i][j]);
if(j>i)
{
k++;
path[k].s=i;
path[k].e=j;
path[k].w=map[i][j];
}
}
printf("%d\n",kruskal(n,k));
}
}
原文地址:http://blog.csdn.net/chaoweilanmao/article/details/27574823