poj 3261 线段树入门

最新推荐文章于 2020-03-29 21:54:06 发布

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Farmer John组织了一场基于牛群高度差异的游戏，通过输入牛群数量、潜在游戏组及每头牛的高度，输出各组中最高与最低牛的高度差。

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 38134		Accepted: 17860
Case Time Limit: 2000MS

Description

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

Sample Output

6
3
0

Source

USACO 2007 January Silver

题目链接：http://poj.org/problem?id=3264

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#define N 50005
using namespace std;
int num[N];
struct Tree
{
    int l;
    int r;
    int maxn;
    int minn;
}tree[N*4];
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    if(tree[root].l==tree[root].r)
    {
        tree[root].maxn=num[l];
        tree[root].minn=num[l];
        return;
    }
    int mid=(r+l)/2;
    build(root*2,l,mid);
    build(root*2+1,mid+1,r);
    tree[root].maxn=max(tree[root<<1].maxn,tree[root*2+1].maxn);
    tree[root].minn=min(tree[root<<1].minn,tree[root*2+1].minn);
}
int query(int root,int L,int R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    return tree[root].maxn;
    int mid=(tree[root].l+tree[root].r)/2;
    int ret=0;
    if(L<=mid)
    ret=max(ret,query(root*2,L,R));
    if(R>mid)
    ret=max(ret,query(root*2+1,L,R));
    return ret;
}
int query1(int root,int L,int R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    return tree[root].minn;
    int mid=(tree[root].l+tree[root].r)/2;
    int ret=100000000;
    if(L<=mid)
    ret=min(ret,query1(root*2,L,R));
    if(R>mid)
    ret=min(ret,query1(root*2+1,L,R));
    return ret;
}
int main()
{
    int n,m,a,b;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        build(1,1,N);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            printf("%d\n",query(1,a,b)-query1(1,a,b));
        }
    }

    //cout << "Hello world!" << endl;
    return 0;
}