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.
给定 nn 个点的坐标，第 ii 个点的坐标为 (xi,yi)(xi,yi)，这 nn 个点编号为 11 到 nn。给定 mm 条边，第 ii 条边连接第 uiui 个点和第 vivi 个点。现在要求你添加一些边，并且能使得任意一点都可以连通其他所有点。求添加的边的总长度的最小值。
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.
第一行两个整数 n,mn,m 代表点数与边数。
接下来 nn 行每行两个整数 xi,yixi,yi 代表第 ii 个点的坐标。
接下来 mm 行每行两个整数 ui,viui,vi 代表第 ii 条边连接第 uiui 个点和第 vivi 个点。
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.
一行一个实数代表添加的边的最小长度，要求保留两位小数，为了避免误差， 请用 6464 位实型变量进行计算。
Sample 1
| Inputcopy | Outputcopy |
|---|---|
4 1 1 1 3 1 2 3 4 3 1 4 | 4.00 |
Hint
数据规模与约定
对于 100%100% 的整数，1≤n,m≤10001≤n,m≤1000，1≤xi,yi≤1061≤xi,yi≤106，1≤ui,vi≤n1≤ui,vi≤n。
说明
Translated by 一只书虫仔。
最小生成树，但要注意要加上m中已经为树的边数
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<math.h>
int n, m, ll = 0,v,u,ss=0;
double min = 0.00;
int b[1001];
struct pppp {
int x, y;
}bb[1001];
struct s {
int x, y;
double k;
};
struct s a[1000001], d[1000001];
//归并排序将其从小到打排序
void digui(int i, int j) {
if (i >= j) {
return;
}
int mid = i + (j - i) / 2;
digui(i, mid);
digui(mid + 1, j);
int r = i, l = mid + 1, h = i;
while (r <= mid && l <= j) {
if (a[r].k < a[l].k) {
d[h].k = a[r].k;
d[h].x = a[r].x;
d[h].y = a[r].y;
h++;
r++;
}
else {
d[h].k = a[l].k;
d[h].x = a[l].x;
d[h].y = a[l].y;
h++;
l++;
}
}
while (r <= mid) {
d[h].k = a[r].k;
d[h].x = a[r].x;
d[h].y = a[r].y;
h++;
r++;
}
while (l <= j) {
d[h].k = a[l].k;
d[h].x = a[l].x;
d[h].y = a[l].y;
h++;
l++;
}
h = i;
while (h <= j) {
a[h].k = d[h].k;
a[h].x = d[h].x;
a[h].y = d[h].y;
h++;
}
}
//并查集
int cha(int i) {
if (b[i] != i) {
return b[i]= cha(b[i]);
}
else {
return b[i];
}
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 1;i <= n;i++) {
b[i] = i;
}
for (int i = 1;i <= n;i++) {
scanf("%d %d", &bb[i].x, &bb[i].y);
}
for (int i = 1;i <= m;i++) {
scanf("%d %d", &u, &v);
if (b[cha(u)] == b[cha(v)])//可能有已经是同一个合集相连
ss++;
b[cha(u)] = b[cha(v)];
}
int e = 0;
for (int i = 1;i <= n;i++) {
for (int j = i + 1;j <= n;j++) {
a[e].x = i;
a[e].y = j;
a[e].k = (double)(sqrt((double)pow(bb[i].x - bb[j].x,2) + (double)pow(bb[i].y - bb[j].y,2)));
e++;
}
}
digui(0, e - 1);
int i = 0;
while (ll <n-m-1+ss&&i<e) {
if (cha(a[i].x) != cha(a[i].y)) {
min += a[i].k;
b[cha(a[i].y)] = b[cha(a[i].x)];
ll++;
}
i++;
}
printf("%.2lf\n", min);
return 0;
}
