HDU 2888 Check Corners【二维RMQ】-优快云博客

本文介绍了一种解决二维区间最大值查询问题的算法——二维RMQ。通过预处理建立查询表，实现快速区间查询，并判断最大值是否位于查询区间的四个角上。

题意：很简单就是求某一区间的最大值是不是在四个角落；

思路：因为矩阵是300*300的所以想到二维RMQ刚好内存也刚好卡；

二维RMQ初始化的时候要注意

当k=0时

ma[i][j][k][l]=max(ma[i][j][k][l-1],ma[i][j+(1<<(l-1))][k][l-1]);

当k≠0时

ma[i][j][k][l]=max(ma[i][j][k-1][l],ma[i+(1<<(k-1))][j][k-1][l]);

询问的话取两者之间的最大值。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741824
#define llinf 4611686018285162540LL
#define eps 1e-8
#define mod 9223372034707292160LL
#define pi acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define drep(i,a,b) for(int i=a;i>=b;i--)
#define mset(x,val) memset(x,val,sizeof(x))
#define mcpy(x,y) memcpy(x,y,sizeof(y))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define mpii(a,b) make_pair(a,b);
#define NN 101010
#define MM 202020
using namespace std;
typedef long long ll;
typedef long double lb;
typedef pair<int,int> pii;
int n,m,q;
int ma[303][303][9][9];
int mp[303][303];

void init(){
    rep(i,1,n) rep(j,1,m) ma[i][j][0][0]=mp[i][j];
    int len1=(int)(log(double(n))/log(2.0));
    int len2=(int)(log(double(m))/log(2.0));
    rep(k,0,len1) rep(l,0,len2) if(k+l){
        rep(i,1,n) rep(j,1,m){
            if(i+(1<<k)-1<=n&&j+(1<<l)-1<=m){
                if(k)  ma[i][j][k][l]=max(ma[i][j][k-1][l],ma[i+(1<<(k-1))][j][k-1][l]);
                else   ma[i][j][k][l]=max(ma[i][j][k][l-1],ma[i][j+(1<<(l-1))][k][l-1]);
            }
        }
    }
}

int rmq(int x1,int y1,int x2,int y2){
    int len1=(int)(log(double(x2-x1+1))/log(2.0));
    int len2=(int)(log(double(y2-y1+1))/log(2.0));
    x2 = x2-(1<<len1)+1;
    y2 = y2-(1<<len2)+1;
    int a=max(ma[x1][y1][len1][len2],ma[x1][y2][len1][len2]);
    int b=max(ma[x2][y1][len1][len2],ma[x2][y2][len1][len2]);
    return max(a,b);
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        rep(i,1,n) rep(j,1,m) scanf("%d",&mp[i][j]);
        init();
        scanf("%d",&q);
        int r1,r2,c2,c1;
        rep(i,1,q){
            scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
            if(r1>r2)swap(r1,r2);
            if(c1>c2)swap(c1,c2);
            int tmp=rmq(r1,c1,r2,c2);
            printf("%d ",tmp);
            if(tmp==mp[r1][c1]||tmp==mp[r1][c2]
               ||tmp==mp[r2][c1]||tmp==mp[r2][c2])   printf("yes\n");
            else printf("no\n");
        }
    }
    return 0;
}