Description
Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.
Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Two space-separated integers: Xi and Yi
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.
Output
* Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.
Sample Input
4 1
1 1
3 1
2 3
4 3
1 4
Sample Output
4.00
这题与常规最小生成树一样。
然而是玄学题。。。
第一次交mle+wa，去网上找了一份交了，ac。
我不服。
改了在交re。
出了变量名都一样啊。
多交了几次，就迷之ac了。
玄不救非，珍爱生命，远离p站。
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=1001;
int n,m,tp,f[maxn];
double ans;
struct node
{int x,y;double val;
}a[maxn*maxn];
struct fm
{double xx,yy;
}p[maxn];
double dis(fm a, fm b)
{return sqrt((a.xx - b.xx)*(a.xx - b.xx) + (a.yy - b.yy) * (a.yy - b.yy));
}
bool cmp(node c,node d)
{return c.val<d.val;
}
int fnd(int x)
{if(f[x]!=x)f[x]=fnd(f[x]);return f[x];
}
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)f[i]=i;for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].xx,&p[i].yy);for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++){++tp;a[tp].x=i;a[tp].y=j;a[tp].val=dis(p[i],p[j]);}int t1,t2;while(m--){scanf("%d%d",&t1,&t2);f[fnd(t1)]=fnd(t2);}sort(a+1,a+tp+1,cmp);for(int i=1;i<=tp;i++)if(fnd(a[i].x)!=fnd(a[i].y)){f[fnd(a[i].y)]=fnd(a[i].x);ans+=a[i].val;}printf("%.2f\n",ans);return 0;
}