SDUT OJ 效率至上（线段树）

效率至上

Time Limit: 5000 ms Memory Limit: 65536 KiB

Submit Statistic

Problem Description

题意很简单,给出一个数目为n的非有序序列,然后有m次查询.对于每次查询输入两个正整数l,r请输出区间[l,r]的最大值与最小值的差值

Input

第一行：输入两个正整数n,m    (1<=n<=50000,  1<=m<=200000  )；

第二行：输入n个整数  大小范围为[1,100000];

接下来的m行,每次两个正整数l,r (1<=l<=r<=n);

Output

输出区间[l,r]最大值与最小值的差值.

Sample Input

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

多次尝试用一个查询函数中得到差值，发现难度比较大，无奈，那就干脆分两步吧

#include <iostream>
#define INF 999999
using namespace std;

struct node
{
    int l, r;
    int Max, Min;
};

int Begin[1000001];
struct node tree[1000001];

void Buildtree( int root, int l, int r )
{
    tree[root].l = l;
    tree[root].r = r;
    if( l == r )
        tree[root].Max = tree[root].Min = Begin[l];
    else
    {
        int mid = ( l + r ) / 2;
        Buildtree( 2 * root + 1, l, mid );
        Buildtree( 2 * root + 2, mid + 1, r);
        tree[root].Max = max( tree[2*root+1].Max, tree[2*root+2].Max );
        tree[root].Min = min( tree[2*root+1].Min, tree[2*root+2].Min );
    }
}

int FindMax ( int root, int l, int r )
{
    int i = tree[root].l, j = tree[root].r;
    if( i > r || j < l )
        return 0;

    l = max( l, i );
    r = min( r, j );
    if( i == l && j == r )
        return tree[root].Max;

    return max( FindMax( 2*root+1, l, r ), FindMax( 2*root+2, l, r ));
}

int FindMin ( int root, int l, int r )
{
    int i = tree[root].l, j = tree[root].r;
    if( i > r || j < l )
        return INF;

    l = max( l, i );
    r = min( r, j );
    if( i == l && j == r )
        return tree[root].Min;

    return min( FindMin( 2*root+1, l, r ), FindMin( 2*root+2, l, r ));
}

int main()
{
    int i, n, m;
    cin >> n >> m;
    for( i=0; i<n; i++ )
        cin >> Begin[i];
    Buildtree ( 0, 0, n-1 );

    while( m-- )
    {
        int l, r;
        cin >> l >> r;
        cout << FindMax ( 0, l-1, r-1 )  -  FindMin ( 0, l-1, r-1 ) << endl;;
    }
    return 0;
}