[USACO16FEB]负载平衡（枚举暴力/模拟）

题目

传送门

题解

这道sb模拟都写不出来，还搞什么线段树，枚举切的边就可以过

代码

#include 
#include 
#include 
#include 
using namespace std;
const int maxn=1e6+100;
const int inf=1e9;
int read(){
    char ch=getchar(); int now=0,f=1;
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0'&&ch<='9') {now=(now<<1)+(now<<3)+ch-'0'; ch=getchar();}
    return now*f;}
struct Cow{
    int x,y;
}a[maxn],b[maxn];
int n,top,ans=inf,head[maxn];

bool cmpx(Cow x,Cow y) {return x.x<y.x;}
bool cmpy(Cow x,Cow y) {return x.y<y.y;}

void work(int k)//k是竖线
{
    int l=0,r=0,L=0,R=0,now=1;
    for (int i=1; i<=n; i++) if (a[i].xfor (int i=1; i<=top; i++)
    {
        int kk=head[i]+1;
        while (now<=n && b[now].y<=kk)
        {
            if (b[now].xelse r++;
            now++;
        }
//      printf("%d %d\n",l,r);
     ans=min(ans,max(max(l,L-l),max(r,R-r)));
 }
//    printf("%d\n",ans);
}

int main()
{
    n=read();
    for (int i=1; i<=n; i++)
    {
        a[i].x=read(); a[i].y=read(); b[i]=a[i];
    }
    sort(a+1,a+1+n,cmpx);
    sort(b+1,b+1+n,cmpy);
    for (int i=1; i<=n; i++) if (b[i].y!=head[top]) head[++top]=b[i].y;//标记有牛出现的行
//  for (int i=1; i<=top; i++) printf("%d ",head[i]);
    for (int i=1; i<=n; i++)
        if (a[i].x!=a[i-1].x) work(a[i].x+1);
    printf("%d\n",ans);
    return 0;
}