POJ-2019——Cornfields(二維RMQ？暴力吧！)

題意：給你一個N*N的矩陣，每一個點有一個權值，詢問子矩陣中最大值與最小值的差

分析：感覺沒啥好說的，這是一個顯然的二維RMQ問題，把模版一套，AC！要注意的就是如果數組開int會爆內存，所以用short把！很奇怪的是這道題很水，暴力也可以過，真是見了鬼了！

二維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;
}