#RMQ#poj 3264 Balanced Lineup

题目

询问一个区间内最大值与最小值的差。


分析

RMQ
f [ i ] [ j ] = m a x ( f [ i ] [ j 1 ] , f [ i + 2 j 1 ) ] [ j 1 ] )


代码

#include <cstdio>
#include <cctype>
using namespace std;
int f[60101][17],f1[60101][17];
int max(int a,int b){return (a>b)?a:b;}
int min(int a,int b){return (a<b)?a:b;}
int in(){
    int ans=0; char c=getchar();
    while (!isdigit(c)) c=getchar();
    while (isdigit(c)) ans=ans*10+c-48,c=getchar();
    return ans;
}
int main(){
    int n,q,x,y,z;
    n=in(); q=in();
    for (int i=1;i<=n;i++) f[i][0]=f1[i][0]=in();
    for (int j=1;(1<<(j-1))<=n;j++)
    for (int i=1;i+(1<<j)-1<=n;i++)
    f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]),
    f1[i][j]=min(f1[i][j-1],f1[i+(1<<(j-1))][j-1]);
    for (int i=1;i<=q;i++){
        x=in(); y=in(); z=0;
        while ((1<<(z+1))<=y-x+1) z++;
        printf("%d\n",max(f[x][z],f[y-(1<<z)+1][z])-min(f1[x][z],f1[y-(1<<z)+1][z]));
    }
    return 0;
} 

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转载自blog.csdn.net/sugar_free_mint/article/details/80752546