HDOJ 1166 HDU 1166 敌兵布阵 ACM 1166 IN HDU

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋    

 

先前做了这一题, 不过是用 树状数组做的, 对于这一类型的题目也来说非常方便快捷. 具体地址 : http://www.cnblogs.com/MiYu/archive/2010/08/25/1808441.html

 

这几天学了 线段树 , 不是很明白它的用途 和 使用方法,  因为 听说 树状数组是 线段树的 一种特殊情况 ( 部分资料是这么说的 ) .  于是就又做了一次这个题目.

不过使用的是线段树的 方法 :

代码如下 :

/*
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
          http://www.cnblog.com/MiYu
Author By : MiYu
Test      : 3
Program   : 1166
*/

#include <iostream>
#include <algorithm>
using namespace std;
typedef struct Line{		//线段树结构体
        friend struct optLine;
        Line () { sum = 0; left = right = NULL; };
        int l,r,sum;
        struct Line *left, *right;
}L;
typedef struct optLine {	//线段树操作结构
       void Creat ( struct Line *node, int beg, int end );
       void modify ( struct Line *node, int pos,int val );
       int quy ( struct Line *node, int beg, int end );     
}OPT;
void optLine::Creat ( struct Line *node, int beg, int end ){	//建树
     node->l = beg;  node->r = end;
     if ( beg == end ) return;
     int mid = ( node->l + node->r ) >> 1;
     L *p = new L;  L *q = new L;
     node->left = p; node->right = q;
     Creat ( p, beg, mid );   Creat ( q, mid + 1, end );          
}
void optLine::modify ( struct Line *node, int pos, int val ){	//修改
     if ( pos >= node->l && pos <= node->r ) node->sum += val;
     if ( node->l == node->r ) return;
     int mid = ( node->l + node->r ) >> 1;
     if ( mid >= pos ) modify ( node->left, pos, val );
     else if ( mid < pos ) modify ( node->right, pos,val );
}
int optLine::quy ( struct Line *node, int beg, int end ){	//查询
    int sum = 0;  int mid = ( node->l + node->r ) >> 1;
    if ( node->l == beg && node->r == end ) return node->sum;
    if ( mid >= end ) return sum += quy ( node->left, beg,end );
    else if ( beg > mid ) return sum += quy ( node->right, beg, end ); 
    else {
          return sum += quy ( node->left, beg, mid ) + quy ( node->right, mid+1,end );     
    }   
}
inline bool scan_ud(int &num) 
{
        char in;
        in=getchar();
        if(in==EOF) return false;
        while(in<'0'||in>'9') in=getchar();
        num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        return true;
}
int main ( int T, int N )
{
    scan_ud ( T );
    for ( int ca = 1; ca <= T ; ++ ca ){
         printf ( "Case %d:\n",ca );
         scan_ud ( N );    OPT opt;  L *root = new L;   int val,x,y;
         opt.Creat ( root, 1, N );  
         for ( int i = 1; i <= N; ++ i ){
              scan_ud ( val );  opt.modify ( root, i, val );    
         }
         char ask[10];  
         while ( scanf ( "%s", ask ), ask[0] != 'E' ){
                scanf ( "%d%d",&x,&y );
                switch ( ask[0] ){
                        case 'Q':  printf ( "%d\n",opt.quy ( root, x,y ) ); break;
                        case 'A':  opt.modify ( root, x, y ); break;
                        case 'S':  opt.modify ( root, x, -y ); break;      
                }      
         }
    }
    return 0;
}