POJ 3264 Balanced Lineup

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题意：给n个数字，有q个查询，每个查询给出两个数l、r，求区间[l,r]内最大值与最小值的差

解题思路：线段树.由于数据量很大，直接模拟一定会超时，所以要用线段树维护每个区间的最大值和最小值，直接套模板

代码：

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>using namespace std;#define INF 0x3f3f3f3fconst int maxn=50000+5;int arr[maxn];struct SegTreeNode{    int valx,vald;}segTree[maxn*4];void build(int root,int istart,int iend){    if(istart==iend)    {        segTree[root].valx=arr[istart];        segTree[root].vald=arr[istart];    }    else    {        int mid=(istart+iend)/2;        build(root*2,istart,mid);        build(root*2+1,mid+1,iend);        segTree[root].valx=min(segTree[root*2].valx,segTree[root*2+1].valx);        segTree[root].vald=max(segTree[root*2].vald,segTree[root*2+1].vald);    }}int query1(int root,int nstart,int nend,int qstart,int qend){    if(qstart>nend||qend<nstart)return INF;    if(qstart<=nstart&&qend>=nend)return segTree[root].valx;    int mid=(nstart+nend)/2;    return min(query1(root*2,nstart,mid,qstart,qend),query1(root*2+1,mid+1,nend,qstart,qend));}int query2(int root,int nstart,int nend,int qstart,int qend){    if(qstart>nend||qend<nstart)return -INF;    if(qstart<=nstart&&qend>=nend)return segTree[root].vald;    int mid=(nstart+nend)/2;    return max(query2(root*2,nstart,mid,qstart,qend),query2(root*2+1,mid+1,nend,qstart,qend));}void updateOne(int root,int nstart,int nend,int index,int addval){    if(nstart==nend)    {        if(index==nstart)        {            segTree[root].valx+=addval;            segTree[root].vald+=addval;            return ;        }    }    int mid=(nstart+nend)/2;    if(index<=mid)updateOne(root*2,nstart,mid,index,addval);    else updateOne(root*2+1,mid+1,nend,index,addval);    segTree[root].valx=min(segTree[root*2].valx,segTree[root*2+1].valx);    segTree[root].vald=max(segTree[root*2].vald,segTree[root*2+1].vald);}int n,q;int main(){    while(scanf("%d%d",&n,&q)==2)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&arr[i]);        }        build(1,1,n);        int l,r;        for(int i=0;i<q;i++)        {            scanf("%d%d",&l,&r);            int mmin=query1(1,1,n,l,r);            int mmax=query2(1,1,n,l,r);            int dif=mmax-mmin;            printf("%d\n",dif);        }    }    return 0;}/*6 31734251 54 62 2*/

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