有一个 a\times b 的整数组成的矩阵,现请你从中找出一个 n\times n 的正方形区域,使得该区域所有数中的最大值和最小值的差最小。
肯定会TLE。 考虑优化。 发现该式子很像RMQ。于是可得:
#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define int long long
using namespace std;
inline int read(){
int res=0,f=1;char ch=getchar();
while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
return res*f;
}
inline void write(int x){
if(x<0) putchar('-'),x=-x;
if(x<10) putchar(x+'0');
else{
write(x/10);
putchar(x%10+'0');
}
}
//queue<int> q;
//set<int> s;
//priority_queue<int> q1;
//priority_queue<int,vector<int>,greater<int> > q2;
//list<int> l;
//stack<int> s;
int a,b,n,g[1010][1010],ans=2e18,logn=20;
int dp[1500][1500];
int dp1[1500][1500];
void st(){
for(int k=0;k<logn;k++){
for(int i=0;i+(1<<k)<a;i++){
for(int j=0;j+(1<<k)<b;j++){
dp[i][j]=max(dp[i][j],max(dp[i+(1<<k)][j+(1<<k)],max(dp[i+(1<<k)][j],dp[i][j+(1<<k)])));
dp1[i][j]=min(dp1[i][j],min(dp1[i+(1<<k)][j+(1<<k)],min(dp1[i+(1<<k)][j],dp1[i][j+(1<<k)])));
}
}
}
}
signed main(){
a=read();b=read();n=read();
for(register int i=0;i<a;i++){
for(register int j=0;j<b;j++) dp[i][j]=dp1[i][j]=g[i][j]=read();
}
logn=log2(n);
st();
for(register int i=0;i<=a-n;i++){
for(register int j=0;j<=b-n;j++){
int Max=0,Min=0;
Max=max(dp[i][j],max(dp[i+n-(1<<logn)][j+n-(1<<logn)],max(dp[i+n-(1<<logn)][j], dp[i][j+n-(1<<logn)])));
Min=min(dp1[i][j],min(dp1[i+n-(1<<logn)][j+n-(1<<logn)],min(dp1[i+n-(1<<logn)][j], dp1[i][j+n-(1<<logn)])));
ans=min(ans,Max-Min);
}
}
write(ans);putchar('\n');
return 0;
}
理论上这道题用二维线段树、二维RMQ都是可以的。 结果我全炸了。。。