poj 3264Balanced Lineup

poj 3264Balanced Lineup

题意：求 一段 区间 的 最大值和最小值 的差值

题解：线段树

碎碎念：某种意义上说，第一道自己手写的线段树，总之蛮好~

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
const int MAXN = 50000+10;
const int INF =  1000000;
int arr[MAXN];
int segx[MAXN<<2];
int segi[MAXN<<2];
void build(int l,int r,int x){
    if(l==r){
        segx[x] = arr[l];
        segi[x] = arr[l];
        return;
    }
    int mid = (l+r)>>1;
    build(l,mid,x<<1);
    build(mid+1,r,x<<1|1);
    segx[x]= max(segx[x<<1],segx[x<<1|1]);
    segi[x]= min(segi[x<<1],segi[x<<1|1]);
    return;
}
int mi,mx;
void quercy(int l,int r,int x,int L,int R){
    if(l>=L && r<=R){
        mi = min(mi,segi[x]);
        mx = max(mx,segx[x]);
        return;
    }
    int mid = (l+r)>>1;
    if(mid>=L){
        quercy(l,mid,x<<1,L,R);
    }
    if(mid<R){
        quercy(mid+1,r,x<<1|1,L,R);
    }
}
int main(){
//    freopen("input.txt","r",stdin);
    int n,q;
    while(~scanf("%d%d",&n,&q)){
        for(int i = 1;i<=n;i++){
            scanf("%d",&arr[i]);
        }
        build(1,n,1);
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            mi = INF;
            mx = 1;
            quercy(1,n,1,l,r);
            printf("%d\n",mx-mi);
        }
    }
    return 0;
}

题解2：平方分割

分桶法：把一排物品或平面分成桶，每个桶维护自己的信息

把一段区间 维护成 一个桶，然后求其最值

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
const int MAXN = 50000+10;
const int INF =  1000000;
int arr[MAXN];
const int sz = 200;
int bucket[260][sz];
int MX[260];
int MI[260];
int main(){
//    freopen("input.txt","r",stdin);
    int n,q;
    while(~scanf("%d%d",&n,&q)){
        fill(MI,MI+260,INF);
        fill(MX,MX+260,1);
        for(int i = 1;i<=n;i++){
            int x = i/sz;
            int y = i%sz;
            scanf("%d",&bucket[x][y]);
            MI[x] = min(MI[x],bucket[x][y]);
            MX[x] = max(MX[x],bucket[x][y]);
        }
        bucket[0][0] = bucket[0][1];
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            int mi = INF;
            int mx = 1;
            int x1 = l/sz;
            int y1= l%sz;
            int x2 = r/sz;
            int y2 = r%sz;
            if(x1==x2){
                for(int i = y1;i<=y2;i++){
                    mi = min(bucket[x1][i],mi);
                    mx = max(bucket[x1][i],mx);
                }
                printf("%d\n",mx-mi);
                continue;
            }
            for(int i = y1;i < sz;i++){
                mi = min(bucket[x1][i],mi);
                mx = max(bucket[x1][i],mx);
            }
            for(int i = 0;i<=y2;i++){
                mi = min(bucket[x2][i],mi);
                mx = max(bucket[x2][i],mx);
            }
            for(int i=x1+1;i<x2;i++){
                mi = min(MI[i],mi);
                mx = max(MX[i],mx);
            }
            printf("%d\n",mx-mi);
        }
    }
    return 0;
}