HDU 2888 Check Corners(简单二维RMQ)

HDU 2888 Check Corners(简单二维RMQ)

http://acm.hdu.edu.cn/showproblem.php?pid=2888

题意：

        给定一个n * m的矩阵，再给定q个询问，每次询问(r1，c1)为左上角，(r2,c2)为右下角的子矩形的最大值，并且判断该最大值是否出现在了这个子矩阵的4个顶角上？

分析:

        基本的二维RMQ应用.

        注意题目中存的矩阵是无符号整数的矩阵.

AC代码:3625ms

<span style="font-size:18px;">#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN =305;
unsigned int val[MAXN][MAXN];
unsigned int dmax[MAXN][MAXN][9][9];
void initRMQ(int n,int m)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            dmax[i][j][0][0] = val[i][j];
    for(int ii=0;(1<<ii)<=n;ii++)
    for(int jj=0;(1<<jj)<=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)
            dmax[i][j][ii][jj]=max(dmax[i][j][ii-1][jj] , dmax[i+(1<<(ii-1))][j][ii-1][jj]);
        else
            dmax[i][j][ii][jj]=max(dmax[i][j][ii][jj-1] , dmax[i][j+(1<<(jj-1))][ii][jj-1]);
}
unsigned int getMax(int x1,int y1,int x2,int y2)
{
    int k1=0;
    while((1<<(k1+1))<=x2-x1+1)k1++;
    int k2=0;
    while((1<<(k2+1))<=y2-y1+1)k2++;
    x2 = x2-(1<<k1)+1;
    y2 = y2-(1<<k2)+1;
    return max(max(dmax[x1][y1][k1][k2] ,dmax[x1][y2][k1][k2] ) , max(dmax[x2][y1][k1][k2] , dmax[x2][y2][k1][k2]));
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2&&n&&m)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%ud",&val[i][j]);
        initRMQ(n,m);
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            unsigned int max_num = getMax(x1,y1,x2,y2);
            bool ok=false;
            if(val[x1][y1]==max_num || val[x1][y2]==max_num || val[x2][y1]==max_num ||val[x2][y2]==max_num)
                ok=true;
            printf("%d %s\n",max_num,ok?"yes":"no");
        }
    }
    return 0;
}
</span>