POJ 3264 Balanced Lineup

#include <functional>  
#include <algorithm>  
#include <iostream>  
#include <fstream>  
#include <sstream>  
#include <iomanip>  
#include <numeric>  
#include <cstring>  
#include <cassert>  
#include <cstdio>  
#include <string>  
#include <vector>  
#include <bitset>  
#include <queue>  
#include <stack>  
#include <cmath>  
#include <ctime>  
#include <list>  
#include <set>  
#include <map>  
using namespace std;
//typedef long long LL;
//typedef __int64 LL; 
//typedef long double DB;  
//typedef unisigned __int64 LL; 
//typedef unsigned long long ULL;
#define EPS  1e-8
#define MAXN 500050
#define INF  0x3f3f3f3f
#define PI   acos(-1.0)
//#define MOD  99991
//#define MOD  99990001 
//#define MOD  1000000007
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define xabs(a) ((a<0)?(-a):a)
#define L(t) (t << 1)  //Left son t*2
#define R(t) (t << 1 | 1) //Right son t*2+1
#define Mid(a,b) ((a+b)>>1) //Get Mid
#define lowbit(a) (a&-a) //Get Lowbit
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
struct node
{
    int l,r,maxval,minval;
} tree[52005*4];

void build(int l,int r,int root)
{
    tree[root].l = l;
	tree[root].r = r;
    tree[root].maxval = 0;
	tree[root].minval = INF;
    if(l == r) return;
    int m = Mid(l,r);
    build(l,m,L(root));
    build(m+1,r,R(root));
}

void modify(int l,int r,int root,int val)
{
	tree[root].maxval=max(val,tree[root].maxval);
	tree[root].minval=min(val,tree[root].minval);
    if(tree[root].l == l && tree[root].r == r)
    {       
        return;
    }
    int m=Mid(tree[root].l,tree[root].r);
    if(r <= m)
		modify(l,r,L(root),val);
    else if(l > m)
		modify(l,r,R(root),val);
    else
    {
        modify(l,m,L(root),val);
        modify(m+1,r,R(root),val);
    }
}
int query(int l,int r,int root)
{
    if(tree[root].l == l && tree[root].r == r)
    {       
        return tree[root].maxval;
    }
    int m=Mid(tree[root].l,tree[root].r);
	if(r <= m)
		return query(l,r,L(root));
    else if(l > m)
		return query(l,r,R(root));
    else
    {
        return max(query(l,m,L(root)),query(m+1,r,R(root)));
    }
}
int query1(int l,int r,int root)
{
    if(tree[root].l == l && tree[root].r == r)
    {       
        return tree[root].minval;
    }
    int m=Mid(tree[root].l,tree[root].r);
	if(r <= m)
		return query1(l,r,L(root));
    else if(l > m)
		return query1(l,r,R(root));
    else
    {
        return min(query1(l,m,L(root)),query1(m+1,r,R(root)));
    }
}
int main()
{
//  freopen("in.txt","r",stdin);
//  freopen("out.txt,"w",stdout);
	int i,n,m,x,y;
    scanf("%d%d",&n,&m);
    build(0,52000,1);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&x);
        modify(i,i,1,x);
    }
	for(i=0;i<m;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		printf("%d\n",xabs(query(a,b,1)-query1(a,b,1)));
	}
	return 0;
}

 

线段树乱搞,就求一个区间最大最小值,每次插入维护区间就行了


 

posted @ 2013-03-24 20:46  xinyuyuanm  阅读(150)  评论(0编辑  收藏  举报