[USACO08OCT]Watering Hole

传送门
Farmer John has decided to bring water to his N (1 <= N <= 300) pastures which are conveniently numbered 1..N. He may bring water to a pasture either by building a well in that pasture or connecting the pasture via a pipe to another pasture which already has water.

Digging a well in pasture i costs W_i (1 <= W_i <= 100,000).

Connecting pastures i and j with a pipe costs P_ij (1 <= P_ij <= 100,000; P_ij = P_ji; P_ii=0).

Determine the minimum amount Farmer John will have to pay to water all of his pastures.

POINTS: 400
输入格式：
第1 行为一个整数n。

第2 到n+1 行每行一个整数，从上到下分别为W_1 到W_n。

第n+2 到2n+1 行为一个矩阵，表示需要的经费（P_ij）。

输出格式：
只有一行，为一个整数，表示所需要的钱数。
输入样例#1：
4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0
输出样例#1：
9

获得最小生成树的新姿势×1

设定一个事实上不存在的点，将所有的点向其连边，边权为在该土地上打井所需的费用，正常做最小生成树，AC~
我才不会告诉你我把并查集写错了呢

CODE：

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;struct aa{    int from,to,dis;}a[100001];int w[301];int c[301][301];int f[305];int n,num,k,x,y,ans;inline void read(int &n){    n=0;char c;bool b=0;    while((c=getchar())==' '||c=='\n'||c=='\r');    if(c=='-') b=1;    else n=c-48;    while(isdigit(c=getchar())) n=n*10+c-48;    if(b) n*=-1;}inline void add(int from,int to,int dis){    a[++num].from=from,a[num].to=to,a[num].dis=dis;}int find(int n){    if(f[n]!=n) f[n]=find(f[n]);    return f[n];}bool cmp(aa a,aa b){    return a.dis<b.dis;}int main(){    read(n);    for(int i=1;i<=n;i++)      read(w[i]);    for(int i=1;i<=n;i++)      for(int j=1;j<=n;j++)        read(c[i][j]);    for(int i=1;i<=n;i++)      f[i]=i;    for(int i=1;i<=n;i++)      for(int j=1;j<i;j++)        add(i,j,c[i][j]);    for(int i=1;i<=n;i++)      add(0,i,w[i]);    sort(a+1,a+num+1,cmp);    for(int i=1;i<=num;i++)    {        x=find(a[i].from),y=find(a[i].to);        if(x!=y)        {            f[y]=x;            ans+=a[i].dis;            k++;        }        if(k==n) break;    }    printf("%d",ans);    return 0;}