Poj 3246 Balanced Lineup（线段树基础）

依旧是线段树基础题

询问区间的最大值和最小值之差，只有询问，没有插入删除。继续理解基础线段树

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 50005
#define INF 0x7FFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;
struct node
{
int l,r,mid,max,min;
}tree[4*MAX];
int hei[MAX],a,b;
int maxx(int a,int b) {
if(a > b) return a;
return b;
}
int minn(int a,int b) {
if(a < b) return a;
return b;
}
void up(int num) {
if(tree[num].max < tree[L(num)].max) tree[num].max = tree[L(num)].max;
if(tree[num].max < tree[R(num)].max) tree[num].max = tree[R(num)].max;
if(tree[num].min > tree[L(num)].min) tree[num].min = tree[L(num)].min;
if(tree[num].min > tree[R(num)].min) tree[num].min = tree[R(num)].min;
}

void build(int l,int r,int num) {
tree[num].l = l;
tree[num].r = r;
tree[num].mid = (l + r) >> 1;
tree[num].max = 0;
tree[num].min = INF;
if(l == r) {
tree[num].max = hei[l];
tree[num].min = hei[l];

return ;
}
build(l,tree[num].mid ,L(num));
build(tree[num].mid+1,r,R(num));

up(num);
}

void query(int l,int r,int num) {
if(l <= tree[num].l && r >= tree[num].r) {
a = maxx(a,tree[num].max);
b = minn(b,tree[num].min);
return ;
}

if(r <= tree[num].mid ) {
query(l,r,L(num));
}
else if(l > tree[num].mid) {
query(l,r,R(num));
}
else {
query(l,tree[num].mid,L(num));
query(tree[num].mid + 1,r,R(num));
}
}

void test(int n)
{
for(int i=1; i<=2*n+1; i++){
printf("l:%d r:%d max:%d min:%d\n",tree[i].l,tree[i].r,tree[i].max,tree[i].min);
}
}

int main() {

int n,q,i;
int l,r;
cin >> n >> q;
for(i=1; i<=n; i++) {
scanf("%d",&hei[i]);
}
build(1,n,1);
//test(n);
for(i=1; i<=q; i++) {
a = 0;
b = INF;
scanf("%d%d",&l,&r);
query(l,r,1);
printf("%d\n",a - b);
}
return 0;
}