Code[VS]1690 开关灯 题解

  1 #include "bits/stdc++.h"
  2 
  3 #define maxN 100010
  4 
  5 using namespace std;
  6 typedef long long QAQ;
  7 
  8 struct Tree
  9 {
 10     int l, r ;
 11     QAQ sum ;//当前开着的灯数量
 12     bool idv;//开关灯的延迟标记
 13 };
 14 
 15 QAQ Min(QAQ a, QAQ b)
 16 {
 17     return a > b ? b : a;
 18 }
 19 
 20 Tree tr[maxN << 2];
 21 
 22 void Push_down ( int i , int m)
 23 {
 24     if(tr[i].idv)//当前结点有延迟标记
 25     {
 26         tr[i << 1].idv = !tr[i << 1].idv ;//左
 27         tr[i << 1 | 1].idv = !tr[i << 1 | 1].idv ;//右
 28         tr[i << 1].sum = tr[i << 1].r - tr[i << 1].l + 1 - tr[i << 1].sum ;//所以灯开关状态取反
 29         tr[i << 1 | 1].sum = tr[i << 1 | 1].r - tr[i << 1 | 1].l + 1 - tr[i << 1 | 1].sum ;//同上
 30         tr[i].idv = !tr[i].idv ;//清零！！！
 31     }
 32 }
 33 
 34 void Build_Tree (int x , int y, int i)
 35 {
 36     tr[i].l = x ;//左端点
 37     tr[i].r = y ;//右端点
 38     if( x == y)return ;
 39     else
 40     {
 41         QAQ mid = (tr[i].l + tr[i].r ) >> 1 ;
 42         Build_Tree ( x , mid , i << 1);//左递归
 43         Build_Tree (mid + 1 , y , i << 1 | 1);//右递归
 44     }
 45 }
 46 
 47 void Update_Tree (int q , int w , int i)
 48 {
 49     if( w >= tr[i].r && q <= tr[i].l)//被完全包含
 50     {
 51         tr[i].idv = !tr[i].idv;//延迟标记
 52         QAQ tot = tr[i].r - tr[i].l + 1;//当前结点总共的灯
 53         tr[i].sum = tot - tr[i].sum;//开关状态全部取反
 54         return ;
 55     }
 56     else
 57     {
 58         Push_down( i , tr[i].r - tr[i].l + 1 );//信息下传函数
 59         QAQ mid = (tr[i].l + tr[i].r) >> 1;
 60         if( q > mid )
 61         {
 62             Update_Tree ( q , w , i << 1 | 1);
 63         }
 64         else if ( w <= mid )
 65         {
 66             Update_Tree ( q , w , i << 1);
 67         }
 68         else
 69         {
 70             Update_Tree ( q , w , i << 1 | 1);
 71             Update_Tree ( q , w , i << 1);
 72         }
 73         tr[i].sum = tr[i << 1].sum + tr[i << 1 | 1].sum ;//回溯更新
 74     }
 75 }
 76 
 77 QAQ Query_Tree (int q , int w , int i )
 78 {
 79     if( w >= tr[i].r && q <= tr[i].l)
 80     {
 81         return tr[i].sum;//被完全包含直接返回值
 82     }
 83     else
 84     {
 85         Push_down( i , tr[i].r - tr[i].l + 1 );
 86         QAQ mid = (tr[i].l + tr[i].r ) >> 1;
 87         if( q > mid )
 88         {
 89             return Query_Tree ( q , w , i << 1 | 1);
 90         }
 91         else if ( w <= mid )
 92         {
 93             return Query_Tree ( q , w , i << 1);
 94         }
 95         else
 96         {
 97             return Query_Tree ( q , w , i << 1 | 1) + Query_Tree ( q , w , i << 1);
 98         }
 99     }
100 }
101 
102 int main ( )
103 {
104     QAQ N, M;
105     int op, l, r;
106     scanf("%d%d", &N, &M);
107     Build_Tree ( 1 , N , 1 );//建树操作
108     while (M--)
109     {
110         scanf("%d", &op);
111         if( !op )
112         {
113             scanf("%d%d", &l, &r);
114             Update_Tree ( l , r , 1 );//更新树
115         }
116         else
117         {
118             scanf("%d%d", &l, &r);
119             printf("%lld\n", Query_Tree ( l , r , 1 ));//查询
120         }
121     }
122     return 0 ;
123 }