poj 3264 Balanced Lineup ST

传送门

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题目大意：
给出n个数，q个询问，求出每个区间中max-min的值

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分析：
ST裸题 ……(⊙o⊙) >_<

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代码如下：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=50005;
int n,m,a[maxn],fmax[maxn][25],fmin[maxn][25];
void ST(){
    for(int j=1;j<=20;j++)
        for(int i=1;i+(1<<(j-1))<=n;i++)
            fmax[i][j]=max(fmax[i][j-1],fmax[i+(1<<(j-1))][j-1]);
    for(int j=1;j<=20;j++)
        for(int i=1;i+(1<<(j-1))<=n;i++)
            fmin[i][j]=min(fmin[i][j-1],fmin[i+(1<<(j-1))][j-1]);
}
int askmax(int x,int y){
    int len=y-x+1;
    int k;
    for(k=20;k>=0;k--)
        if(len&(1<<k)||k==0)    
            break;
    return max(fmax[x][k],fmax[y-(1<<k)+1][k]);
}
int askmin(int x,int y){
    int len=y-x+1;
    int k;
    for(k=20;k>=0;k--)
        if(len&(1<<k)||k==0)    
            break;
    return min(fmin[x][k],fmin[y-(1<<k)+1][k]);
}
signed main(void){
    scanf("%d%d",&n,&m);
    memset(fmax,-inf,sizeof(fmax));
    memset(fmin,inf,sizeof(fmin));
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),fmax[i][0]=fmin[i][0]=a[i];
    ST();
    while(m--){
        int x,y;
        scanf("%d%d",&x,&y);
        printf("%d\n",askmax(x,y)-askmin(x,y));
    }
    return 0;
}


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感谢yousiki


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by >o< neighthorn