noi.ac#36 题解

标签：get   cpp   const   预处理   getchar   put   span   sum   http   

题目链接

\(Solution:\)

直接大力预处理,\(O(n^2log_2n)+O(q)\)可过

\(Code:\)

#include<bits/stdc++.h>
using namespace std;
namespace my_std
{
    typedef long long ll;
    typedef double db;
    const db PI=acos(-1.0);
    #define cp complex<db>
    #define MP make_pair
    #define fir first
    #define sec second
    #define fr(i,x,y) for(int i=(x);i<=(y);i++)
    #define pfr(i,x,y) for(int i=(y);i>=(x);i--)
    #define gfr(u) for(int i=head[u];i;i=e[i].nxt)
    #define pf printf
    inline ll read()
    {
        ll sum=0,f=1;
        char ch=0;
        while(!isdigit(ch))
        {
            if(ch=='-') f=-1;
            ch=getchar();
        }
        while(isdigit(ch))
        {
            sum=(sum<<1)+(sum<<3)+(ch^48);
            ch=getchar();
        }
        return sum*f;
    }
    inline void write(int x)
    {
        if(x<0)
        {
            putchar('-');
            x=-x;
        }
        if(x>9) write(x/10);
        putchar(x%10+'0');
    }
}
using namespace my_std;
const int N=1050;
int n,m,q,a[N][N],b[N][N],aa[N][N],bb[N][N],ans[N][N]; 
int main(void)
{
    n=read(),m=read(),q=read();
    fr(i,1,n) fr(j,1,m) a[i][j]=read();
    fr(i,1,m) fr(j,1,n) b[i][j]=a[j][i];
    fr(i,1,n) fr(j,1,m) aa[i][j]=a[i][j];
    fr(i,1,m) fr(j,1,n) bb[i][j]=b[i][j];
    fr(i,1,n) sort(aa[i]+1,aa[i]+m+1);
    fr(i,1,m) sort(bb[i]+1,bb[i]+n+1);
    fr(i,1,n) 
    {
        fr(j,1,m)
        {
            int x=lower_bound(aa[i]+1,aa[i]+m+1,a[i][j])-aa[i];
            int y=lower_bound(bb[j]+1,bb[j]+n+1,a[i][j])-bb[j];
            ans[m-x+1][n-y+1] ++;
        }
    }
    while(q--)
    {
        int x=read(),y=read();
        write(ans[x][y]);
        putchar('\n');
    }
    return 0;
}

noi.ac#36 题解

标签：get   cpp   const   预处理   getchar   put   span   sum   http   

原文地址：https://www.cnblogs.com/lgj-lgj/p/12335545.html