二维RMQ_Cornfields SCU - 2154

这个题scu的测试比较好

用printf,注意数组的大小

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he's looking to build the cornfield on the flattest piece of land he can find.

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it.

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form "in this B x B submatrix, what is the maximum and minimum elevation?". The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield.

Input * Line 1: Three space-separated integers: N, B, and K.

* Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc.

* Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1.
Output * Lines 1..K: A single integer per line representing the difference between the max and the min in each query.
Sample Input

5 3 1
5 1 2 6 3
1 3 5 2 7
7 2 4 6 1
9 9 8 6 5
0 6 9 3 9
1 2

Sample Output

5

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) begin(x),end(x)
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}


//二维RMQ，预处理复杂度 n*m*log*(n)*log(m)
//数组下标从 1 开始

const int maxn=256;
const int max_log=9;
int val[maxn][maxn];
int dp[maxn][maxn][max_log][max_log];//最大值
int dp2[maxn][maxn][max_log][max_log];
int mm[maxn]; //二进制位数减一
void initRMQ(int n,int m)
{
    mm[0] = -1;
    for(int i = 1; i <= 305; i++)
        mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++){
            dp[i][j][0][0] = val[i][j];
            dp2[i][j][0][0] = val[i][j];
        }
    for(int ii = 0; ii <= mm[n]; ii++)
        for(int jj = 0; jj <= mm[m]; jj++)
            if(ii+jj)
                for(int i = 1; i + (1<<ii) - 1 <= n; i++)
                for(int j = 1; j + (1<<jj) - 1 <= m; j++)
                {
                    if(ii){
                        dp[i][j][ii][jj]= max(dp[i][j][ii-1][jj],dp[i+(1<<(ii-1))][j][ii-1][jj]);
                        dp2[i][j][ii][jj]= min(dp2[i][j][ii-1][jj],dp2[i+(1<<(ii-1))][j][ii-1][jj]);
                    }
                    else{
                        dp[i][j][ii][jj]= max(dp[i][j][ii][jj-1],dp[i][j+(1<<(jj-1))][ii][jj-1]);
                        dp2[i][j][ii][jj]= min(dp2[i][j][ii][jj-1],dp2[i][j+(1<<(jj-1))][ii][jj-1]);
                    }
                }
}
//get Min x1<=x2,y1<=y2)
int rmq(int x1,int y1,int x2,int y2)
{
    int k1 = mm[x2-x1+1];
    int k2 = mm[y2-y1+1];
    x2 = x2 - (1<<k1) + 1;
    y2 = y2 - (1<<k2) + 1;
    return max(max(dp[x1][y1][k1][k2],
                dp[x1][y2][k1][k2])
               ,max(dp[x2][y1][k1][k2]
                ,dp[x2][y2][k1][k2]));
}

int rmq2(int x1,int y1,int x2,int y2)
{
    int k1 = mm[x2-x1+1];
    int k2 = mm[y2-y1+1];
    x2 = x2 - (1<<k1) + 1;
    y2 = y2 - (1<<k2) + 1;
    return min(min(dp2[x1][y1][k1][k2],
                dp2[x1][y2][k1][k2])
               ,min(dp2[x2][y1][k1][k2]
                ,dp2[x2][y2][k1][k2]));
}

int main(){
   // freopen("in.txt","r",stdin);
   int n,b,k;
    while(~scanf("%d%d%d",&n,&b,&k)){
        b--;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        val[i][j]=rd();
        initRMQ(n,n);
        /*
        for(int i=0;i<10;i++)
            cout<<mm[i]<<" ";
        cout<<endl;
        */
        while(k--){
            int x=rd(),y=rd();
            int a=rmq(x,y,min(x+b,n),min(y+b,n));
            int c=rmq2(x,y,min(x+b,n),min(y+b,n));
            printf("%d\n",a-c);
        }
    }
}