POJ-2019——Cornfields(二维RMQ？暴力吧！)
题意：给你一个N*N的矩阵，每一个点有一个权值，询问子矩阵中最大值与最小值的差
分析：感觉没啥好说的，这是一个显然的二维RMQ问题，把模版一套，AC！要注意的就是如果数组开int会爆内存，所以用short把！很奇怪的是这道题很水，暴力也可以过，真是见了鬼了！
二维ST算法
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 250+5
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define IT iterator
typedef long long ll;
const double eps = 1e-9;
const double pi = acos(-1);
const ll mod = 1e9+7;
short num[maxn][maxn];
short maxx[maxn][maxn][8][8];
short minn[maxn][maxn][8][8];
void rmq(int n,int m)
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
maxx[i][j][0][0]=minn[i][j][0][0]=num[i][j];
for(int i=0;i<8;i++)
for(int j=0;j<8;j++)
{
if(i==0&&j==0)continue;
for(int p=1;p+(1<<i)-1<=n;p++)
for(int q=1;q+(1<<j)-1<=n;q++)
if(i==0)
{
maxx[p][q][i][j]=max(maxx[p][q][i][j-1],maxx[p][q+(1<<j-1)][i][j-1]);
minn[p][q][i][j]=min(minn[p][q][i][j-1],minn[p][q+(1<<j-1)][i][j-1]);
}
else
{
maxx[p][q][i][j]=max(maxx[p][q][i-1][j],maxx[p+(1<<i-1)][q][i-1][j]);
minn[p][q][i][j]=min(minn[p][q][i-1][j],minn[p+(1<<i-1)][q][i-1][j]);
}
}
}
int query(int a,int b,int p,int q)
{
int kr=(int)(log(p-a+1.0)/log(2.0));
int kc=(int)(log(q-b+1.0)/log(2.0));
short mi=minn[a][b][kr][kc];
mi=min(mi,minn[p-(1<<kr)+1][b][kr][kc]);
mi=min(mi,minn[a][q-(1<<kc)+1][kr][kc]);
mi=min(mi,minn[p-(1<<kr)+1][q-(1<<kc)+1][kr][kc]);
short ma=maxx[a][b][kr][kc];
ma=max(ma,maxx[p-(1<<kr)+1][b][kr][kc]);
ma=max(ma,maxx[a][q-(1<<kc)+1][kr][kc]);
ma=max(ma,maxx[p-(1<<kr)+1][q-(1<<kc)+1][kr][kc]);
return ma-mi;
}
int main()
{
int n,b,k;
while(~scanf("%d %d %d",&n,&b,&k))
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&num[i][j]);
rmq(n,n);
while(k--)
{
int p,q;
scanf("%d %d",&p,&q);
printf("%d\n",query(p,q,p+b-1,q+b-1));
}
}
return 0;
}
暴力
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 250+5
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define IT iterator
typedef long long ll;
const double eps = 1e-9;
const double pi = acos(-1);
const ll mod = 1e9+7;
int num[maxn][maxn];
int main()
{
int n,b,k;
while(~scanf("%d %d %d",&n,&b,&k))
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&num[i][j]);
while(k--)
{
int minn=251,maxx=-1;
int p,q;
scanf("%d %d",&p,&q);
for(int i=p;i<p+b;i++)
for(int j=q;j<q+b;j++)
{
minn=min(minn,num[i][j]);
maxx=max(maxx,num[i][j]);
}
printf("%d\n",maxx-minn);
}
}
return 0;
}