题意：给出n个点的坐标,要把n个点连通，使得总距离最小，可是有m对点已经连接，输入m，和m组a和b，表示a和b两点已经连接。
思路：两种做法。（1）用prim算法时，输入a，b。令mp[a][b]=0。然后进行一遍prim（2）Kruskal算法+并查集
代码：
//prim写法
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;
const int N = 1005;
struct Node
{
double x,y;
}node[maxn];
int n,m;
double mp[maxn][maxn];
double dist[maxn];
bool vis[maxn];
double Dis(Node n1,Node n2)
{
return sqrt((n1.x-n2.x)*(n1.x-n2.x)+(n1.y-n2.y)*(n1.y-n2.y));
}
double prim()
{
int i,j,now;
double mi;
mem(vis,false);
mem(dist,INF);
for (i=1;i<=n;i++)
dist[i]=mp[1][i];
dist[1]=0;
vis[1]=true;
for (i=1;i<=n;i++)
{
now=-1;
mi=INF;
for (j=1;j<=n;j++)
{
if (!vis[j]&&mi>dist[j])
{
now=j;
mi=dist[j];
}
}
if (now==-1) break;
vis[now]=true;
for (j=1;j<=n;j++)
{
if (!vis[j]&&dist[j]>mp[now][j])
dist[j]=mp[now][j];
}
}
double ans=0;
for (i=1;i<=n;i++)
ans+=dist[i];
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v;
while (~sf(n))
{
for (i=1;i<=n;i++)
scanf("%lf%lf",&node[i].x,&node[i].y);
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
if (i==j) mp[i][j]=0;
else mp[i][j]=Dis(node[i],node[j]);
}
sf(m);
for (i=0;i<m;i++) //在同一个联通块的距离直接赋为0
{
sff(u,v);
mp[u][v]=mp[v][u]=0;
}
printf("%.2f\n",prim());
}
return 0;
}
//Kruskal+并查集写法
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 1000000;
const int MAXM = 200010;
const int N = 1005;
struct Node
{
double x,y;
}node[maxn];
struct Edge
{
int u,v;
double len;
}edge[MAXN];
int n,m,cnt;
int father[maxn];
int cmp(Edge e1,Edge e2)
{
return e1.len<e2.len;
}
void init()
{
cnt=0;
for (int i=0;i<=n;i++)
father[i]=i;
}
double Dis(Node n1,Node n2)
{
return sqrt((n1.x-n2.x)*(n1.x-n2.x)+(n1.y-n2.y)*(n1.y-n2.y));
}
int find_father(int x)
{
if (x!=father[x])
father[x]=find_father(father[x]);
return father[x];
}
double Kruskal()
{
int i,j;
double ans=0;
sort(edge,edge+cnt,cmp);
for (i=0;i<cnt;i++)
{
int fu=find_father(edge[i].u);
int fv=find_father(edge[i].v);
if (fu!=fv)
{
ans+=edge[i].len;
father[fu]=fv;
}
}
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v;
while (~sf(n))
{
init();
for (i=1;i<=n;i++)
scanf("%lf%lf",&node[i].x,&node[i].y);
for (i=1;i<n;i++)
{
for (j=i+1;j<=n;j++)
{
if (i==j) continue;
else
{
double x=Dis(node[i],node[j]);
edge[cnt].u=i;
edge[cnt].v=j;
edge[cnt++].len=x;
}
}
}
sf(m);
for (i=1;i<=m;i++)
{
sff(u,v);
int fu=find_father(u);
int fv=find_father(v);
if (fu!=fv)
father[fu]=fv;
}
printf("%.2f\n",Kruskal());
}
return 0;
}