poj 3264 Balanced Lineup（线段树）

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题目链接：http://poj.org/problem?id=3264

思路：

线段树，维护最大值最小值之差。

一开始蠢得要死，写了两个Query维护最大最小值，后来在网上参考大神的优化了一下。

#include <limits.h>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#include <algorithm>#include <iostream>#include <iterator>#include <sstream>#include <queue>#include <stack>#include <string>#include <vector>#include <map>#include <set>//#define ONLINE_JUDGE#define eps 1e-6#define INF 0x7fffffff                                          //INT_MAX#define inf 0x3f3f3f3f                                          //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++)                          //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define MEM3(a) memset(a,0x3f,sizeof(a))#define MEMS(a) memset(a,'\0',sizeof(a))#define LL __int64using namespace std;const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }template<class T>T Mint(T a, T b, T c) {    if (a>b) {        if (c>b)            return b;        return c;    }    if (c>a)        return a;    return c;}template<class T>T Maxt(T a, T b, T c) {    if (a>b) {        if (c>a)            return c;        return a;    }    else if (c > b)        return c;    return b;} const int maxn=200010;int T,n,m;int maxnum,minnum;int a[maxn]; struct segTree{    int l,r;    int v,u;}node[maxn*3+10]; void build(int id,int left,int right){    node[id].l=left;    node[id].r=right;    node[id].v=node[id].u=0;    if(left==right){        node[id].v=node[id].u=a[left];        return ;    }    int mid=(left+right)/2;    build(id<<1,left,mid);    build(id<<1|1,mid+1,right);    node[id].v=max(node[id<<1].v,node[id<<1|1].v);    node[id].u=min(node[id<<1].u,node[id<<1|1].u);} void Query(int id,int Qleft,int Qright){    int l=node[id].l,r=node[id].r;    if(node[id].v<=maxnum&&node[id].u>=minnum) return ;    if(l>=Qleft&&r<=Qright){        maxnum=max(maxnum,node[id].v);        minnum=min(minnum,node[id].u);        return ;    }    int mid=(l+r)/2;    if(mid>=Qright)        Query(id<<1,Qleft,Qright);    else if(mid<Qleft)        Query(id<<1|1,Qleft,Qright);    else{        Query(id<<1,Qleft,mid);        Query(id<<1|1,mid+1,Qright);    }} int main(){    while(~sfd(n,m)){        for2(i,1,n) sf(a[i]);        build(1,1,n);        int p,q;        while(m--){            sfd(p,q);            minnum=inf,maxnum=-inf;            Query(1,p,q);            pf(maxnum-minnum);        }    }    return 0;}

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