zoukankan html css js c++ java

POJ 3264 Balanced Lineup

题目描述：跟N个数和Q个询问，求询问区间[a, b]中最大值和最小值的差。

我的第一个线段树……

  1 /*
  2 Interval Tree
  3 */
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 
  8 const int MAXN = 50000 + 2;
  9 const int INF = 2147483645;
 10 
 11 struct Node
 12 {
 13     int l, r;
 14     int maxi, mini;
 15     int l_child, r_child;
 16 };
 17 
 18 int cnt;
 19 Node Tree[ MAXN << 2 ];
 20 int findMax, findMin;
 21 
 22 void Build( int l, int r, Node &Td )
 23 {
 24     Td.l = l;
 25     Td.r = r;
 26 
 27     Td.maxi = -INF;
 28     Td.mini = INF;
 29 
 30     if ( l != r )
 31     {
 32         ++cnt;
 33         Td.l_child = cnt;
 34         Build( l, (l + r) / 2, Tree[ Td.l_child ] );
 35 
 36         ++cnt;
 37         Td.r_child = cnt;
 38         Build( (l + r) / 2 + 1, r, Tree[ Td.r_child ] );
 39     }
 40 
 41     return;
 42 }
 43 
 44 void Insert( int ii, int &e, Node &Td )
 45 {
 46     if ( ii == Td.l && ii == Td.r )
 47     {
 48         Td.maxi = Td.mini = e;
 49         return;
 50     }
 51 
 52     Td.maxi = Td.maxi > e ? Td.maxi : e;
 53     Td.mini = Td.mini < e ? Td.mini : e;
 54 
 55     if ( ii <= (Td.l + Td.r) / 2 )
 56         Insert( ii, e, Tree[ Td.l_child ] );
 57     else
 58         Insert( ii, e, Tree[ Td.r_child ] );
 59 
 60     return;
 61 }
 62 
 63 /*
 64 void show( int n )
 65 {
 66     for ( int i = 0; i <= n * 2; ++i )
 67       printf("[%d, %d]: %d %d\n", Tree[i].l, Tree[i].r, Tree[i].maxi, Tree[i].mini );
 68     return;
 69 }
 70 */
 71 
 72 void Query( int st, int ed, Node &Td )
 73 {
 74     if ( Td.maxi <= findMax && Td.mini >= findMin ) return;
 75 
 76     if ( st == Td.l && ed == Td.r )
 77     {
 78         findMax = findMax > Td.maxi ? findMax : Td.maxi;
 79         findMin = findMin < Td.mini ? findMin : Td.mini;
 80         return;
 81     }
 82 
 83     int mid = (Td.l + Td.r) / 2;
 84 
 85     if ( ed <= mid )
 86     {
 87         Query( st, ed, Tree[ Td.l_child ] );
 88     }
 89     else if ( st > mid  )
 90     {
 91         Query( st, ed, Tree[ Td.r_child ] );
 92     }
 93     else
 94     {
 95         Query( st, mid, Tree[ Td.l_child ] );
 96         Query( mid + 1, ed, Tree[ Td.r_child ] );
 97     }
 98 
 99 }
100 
101 int main()
102 {
103     int n, Q;
104     while ( scanf( "%d%d", &n, &Q ) != EOF )
105     {
106         cnt = 0;
107         Build( 1, n, Tree[0] );
108 
109         for ( int i = 0; i < n; ++i )
110         {
111             int a;
112             scanf( "%d", &a );
113             Insert( i + 1, a, Tree[0] );
114         }
115 
116       //  show(n);
117 
118         for ( int i = 0; i < Q; ++i )
119         {
120             int a, b;
121             scanf( "%d%d", &a, &b );
122             findMax = -INF;
123             findMin = INF;
124             Query( a, b, Tree[0] );
125             printf( "%d\n", findMax - findMin );
126         }
127     }
128     return 0;
129 }

查看全文

相关阅读:
菜鸟笔记 -- Chapter 6.2.6 内部类
 菜鸟笔记 -- Chapter 6 面向对象
 菜鸟笔记 -- Chapter 6.1 面向对象概述
 菜鸟笔记 -- Chapter 4.7 代码注释与编码规范
 菜鸟笔记 -- Chapter 4 Java语言基础
 小白袍 -- Chapter 1.4.1.1 URL编码的理论解读
 小白袍 -- Chapter 1.1 避不开的编解码
 小白袍 -- Chapter 1 Java中的Encode与Decode
菜鸟崛起 Ajax
软件测试

原文地址：https://www.cnblogs.com/GBRgbr/p/2734649.html