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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.
<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>