POJ-3264-Balanced Lineup

POJ-3264-Balanced Lineup

http://poj.org/problem?id=3264

线段树，求区间的最大值与最小值之差

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define N 50005
#define INF 0x7ffffff
int num[N];
struct cam
{
	int x;  //起点
	int y;  //终点
	int min; //最小值
	int max; //最大值
}list[N*4];
int Min(int x,int y)
{
	return x<y?x:y;
}
int Max(int x,int y)
{
	return x>y?x:y;
}
void build(int k,int x,int y)
{
	int mid;
	list[k].x=x;
	list[k].y=y;
	if(list[k].x==list[k].y)
	{
		list[k].min=num[x];
		list[k].max=num[x];
		return;
	}
    mid=(x+y)/2;
	build(k<<1,x,mid);
	build(k<<1|1,mid+1,y);
	list[k].min=Min(list[k<<1].min,list[k<<1|1].min);
	list[k].max=Max(list[k<<1].max,list[k<<1|1].max);
    return;
}
int findmin(int k,int x,int y)
{
	int mid;
	if(list[k].x==x&&list[k].y==y)
	return list[k].min;
    mid=(list[k].x+list[k].y)/2;
	if(x>mid)
	return findmin(k<<1|1,x,y);
	else if(y<=mid)
	return findmin(k<<1,x,y);
	return Min(findmin(k<<1,x,mid),findmin(k<<1|1,mid+1,y));
}
int findmax(int k,int x,int y)
{
	int mid;
	if(list[k].x==x&&list[k].y==y)
	return list[k].max;
	mid=(list[k].x+list[k].y)/2;
	if(x>mid)
	return findmax(k<<1|1,x,y);
	else if(y<=mid)
	return findmax(k<<1,x,y);
	return Max(findmax(k<<1,x,mid),findmax(k<<1|1,mid+1,y));
}
int main()
{
	int i,n,m;
	int x,y;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
	scanf("%d",&num[i]);
	build(1,1,n);
	while(m--)
	{
		scanf("%d%d",&x,&y);
		printf("%d\n",findmax(1,x,y)-findmin(1,x,y));
	}
	return 0;
}