题目上仅仅给的坐标，没有给出来边的长度，不管是prim算法还是kruskal算法我们都须要知道边的长度来操作。
这道题是浮点数，也没啥大的差别，处理一下就能够了。
有关这两个算法的介绍前面我已经写过了。就不在多写了
prim算法：
<span style="font-family:Courier New;font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<algorithm>
#include<climits>
using namespace std;
struct node
{
double i,j;
}g[105];
double gra[105][105];
double dist(double a,double b,double c,double d)
{
return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}
int n,cnt=0,T;
void prim()
{
int visit[105],now,i,j;
double dis[105];
double Min;
memset(visit,0,sizeof(visit));
for(i=1; i<=n; i++)
dis[i] = INT_MAX;
visit[1] = 1, dis[1] = 0, now = 1;//now都是当前新加的点
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
if(!visit[j] && dis[j]>gra[now][j])//用新加的点来更新其它点到此集合的距离
dis[j] = gra[now][j];
}
Min = INT_MAX;
for(j=1; j<=n; j++)
{
if(!visit[j] && dis[j] < Min)//每次都找到距离最小的点。加进去
Min = dis[now = j];
}
visit[now] = 1;
}
double sum = 0;
for(i=1; i<=n; i++)
{
sum += dis[i];
}
printf("%.2lf\n",sum);
if(cnt!= T)//注意每两个输出案例之间都有一个换行
cout << endl;
}
int main()
{
int i,j;
cin >> T;
while(cin >> n)
{
cnt ++;
for(i=1; i<=n; i++)
{
cin >> g[i].i >> g[i].j;
}
memset(gra,0,sizeof(gra));
for(i=1; i<=n; i++)
{
for(j=i+1; j<=n; j++)
{
gra[i][j] = gra[j][i] = dist(g[i].i,g[i].j,g[j].i,g[j].j);
}
}
prim();
}
return 0;
}
</span>kruskal算法:
<span style="font-family:Courier New;font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<algorithm>
#include<climits>
using namespace std;
struct node
{
int i,j;
double len;
}gra[10005];
struct node1
{
double i,j;
}g[105];
int p[105];
double dist(double a,double b,double c,double d)
{
return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}
int cmp(const void *a,const void *b)
{
return (((node *)a)->len - ((node *)b)->len > 0) ? 1:-1;
}
int n,cnt=0,T,k;
int find(int x)
{
return x == p[x]? x: p[x] = find(p[x]);
}
void kruskal()
{
double sum = 0;
int i;
for(i=1; i<k; i++)
{
int x = find(gra[i].i);
int y = find(gra[i].j);
if(x!=y)
{
sum += gra[i].len;
p[x] = y;
}
}
printf("%.2f\n",sum);
if(cnt != T)
cout << endl;
}
int main()
{
int i,j;
cin >> T;
while(cin >> n)
{
cnt ++;
for(i=1; i<=n; i++)
{
cin >> g[i].i >> g[i].j;
}
memset(gra,0,sizeof(gra));
k=1;
for(i=1; i<=n; i++)
{
for(j=i+1; j<=n; j++)
{
gra[k].len = dist(g[i].i,g[i].j,g[j].i,g[j].j);
gra[k].i = i;
gra[k].j = j;
k++;
}
}
//prim();
for(i=1; i<=n; i++)
p[i] = i;
qsort(gra+1,k-1,sizeof(gra[0]),cmp);
kruskal();
}
return 0;
}
</span>
