效率至上
Time Limit: 5000 ms Memory Limit: 65536 KiB
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;
}