Luogu 1194买礼物

题意:买n个东西,这些东西有个原价m,若买东西i,则东西j降到a[i][j]元,求买n个东西最少的钱.

做法:若买i后买j的价钱a[i][j]比m小,那么连一条无向边，跑一波最小生成树.

看剩下还有几个点没连进去的就直接按原价买..

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#define ll long long 
#define out(a) printf("%d",a)
#define ln printf("
")
const int N=3e5+50050;
const int MOD=1e9+7;
using namespace std;
int n,m,x;
int tot=0,ans=0,cnt=0;
int pre[N];
struct node
{
    int x,y,cost;
}a[N];
int read()
{
    int s=0,t=1; char c;
    while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
    while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
    return s*t;
}
ll readl()
{
    ll s=0,t=1; char c;
    while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
    while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
    return s*t;
}
bool cmp(node a,node b)
{
    return a.cost<b.cost;
}
int find(int x)
{
    if (pre[x]==x) return x;
    else return pre[x]=find(pre[x]);
}
void Kruskal()
{
    for (int i=1;i<=tot;i++){
      int x=find(a[i].x),y=find(a[i].y);
      if (x!=y){
        ans+=a[i].cost;
        pre[x]=y; cnt++;
      }
      if (cnt==n-1) return;
    }
}
int main()
{
    m=read(),n=read();
    for (int i=1;i<=n;i++){
      pre[i]=i;
      for (int j=1;j<=n;j++){
        x=read();
        if (x!=0&&x<m) a[++tot].x=i,a[tot].y=j,a[tot].cost=x;
      }
    }
    sort(a+1,a+tot+1,cmp);
    Kruskal();
    ans+=(n-cnt)*m;
    out(ans);
    return 0;
}