POJ3264（分桶法）

题目思路直接，用分桶法管理每一块的最大值和最小值就可以

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<map>
#define DEBUG(x) cout << #x << " = " << x << endl
using namespace std;
const int N=50000+10;
const int S=250;
int a[N];
vector<int>vec[S];
int main()
{
    ///freopen("in.txt","r",stdin);
    int n,q;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        vec[i/(S+1)].push_back(a[i]);
    }
    for(int i=0;i<n/(S+1);i++){
        sort(vec[i].begin(),vec[i].end());
    }
    while(q--){
        int L,R;
        scanf("%d%d",&L,&R);
        int Min=0x3f3f3f3f,Max=-1;
        int tl=L,tr=R;
        while(tl<=tr&&tl/(S+1)==L/(S+1)){
            Min=min(a[tl],Min);
            Max=max(a[tl++],Max);
        }
        while(tl<=tr&&tr/(S+1)==R/(S+1)){
            Min=min(a[tr],Min);
            Max=max(a[tr--],Max);
        }
        while(tl<=tr){
            int bn=tl/(S+1);
            int s=0,e=vec[bn].size()-1;
            Min=min(vec[bn][s],Min);
            Max=max(vec[bn][e],Max);
            tl+=S;
        }
        printf("%d\n",Max-Min);
    }
    return 0;
}