HDU 5023 A Corrupt Mayor's Performance Art

HDU 5023A Corrupt Mayor‘s Performance Art （线段树 + 状态压缩）

上周网络赛的B题，题目很长但是前面根本没有用题意：线段树操作

P l r c      将 [l,r] 区间颜色换为 cQ l r　      查询 [l,r]区间一共有多少种颜色，并按升序输出[ 初始状态墙壁颜色全部为2，总共30种颜色 ]

分析①: 考虑到最多30种颜色，可以直接用一个int型的变量来存放状态，32位正好具体就是用线段树的成段更新，lazy为懒惰标记。注意下Query

  1 /* ***********************************************
  2 Problem       :
  3 File Name     :
  4 Author        :
  5 Created Time  :
  6 ************************************************ */
  7 //#define BUG
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <algorithm>
 11 #include <map>
 12 #include <vector>
 13 #include <list>
 14 #include <queue>
 15 #include <ctime>
 16 #include <iostream>
 17 #include <cmath>
 18 #include <set>
 19 #include <string>
 20 using namespace std;
 21 typedef long long LL;
 22 typedef unsigned long long ULL;
 23 #define INF                 0x7fffffff
 24 #define MAX                 0x3f3f3f3f
 25
 26 #define CLR( a, b )         memset( a, b, sizeof(a) )
 27 #define REP( i, a, b )      for( int i = (a); i < (b); ++i )
 28 #define FOR( i, a, b )      for( int i = (a); i <=(b); ++i )
 29 #define FOD( i, a, b )      for( int i = (a); i >=(b); --i )
 30 #define MID                 ( l + r ) >> 1
 31 #define lson                l, m, o << 1
 32 #define rson                m + 1, r, o << 1 | 1
 33 #define ls                  o << 1
 34 #define rs                  o << 1 | 1
 35
 36 #define MAXN                1010101
 37
 38 int lazy[MAXN << 2];
 39 int col[MAXN << 2];
 40
 41 void PushUp( int o )
 42 {
 43     col[o] = col[ls] | col[rs];
 44 }
 45
 46 void PushDown( int o )
 47 {
 48     if( lazy[o] )
 49     {
 50         lazy[ls] = lazy[o];
 51         lazy[rs] = lazy[o];
 52         col[ls] = col[rs] = lazy[o];//向下传递并直接修改col的值
 53         lazy[o] = 0;
 54     }
 55 }
 56
 57 void Build( int l, int r, int o )
 58 {
 59     if( l == r )
 60     {
 61         col[o] = 1 << (2-1);
 62         return;
 63     }
 64     int m = MID;
 65     Build( lson );
 66     Build( rson );
 67     PushUp( o );
 68 }
 69
 70 void Update( int L, int R, int v, int l, int r, int o )
 71 {
 72     if( L <= l && r <= R )
 73     {
 74         lazy[o] = 1 << v;
 75         col[o] = 1 << v;
 76         return;
 77     }
 78     int m = MID;
 79     PushDown( o );
 80     if(L <= m)  Update( L, R, v, lson );
 81     if(R >  m)  Update( L, R, v, rson );
 82     PushUp( o );
 83
 84 }
 85
 86 int Query ( int L , int R , int l , int r , int o )
 87 {
 88     if( L <= l && r <= R ) return col[o];
 89     int m = MID;
 90     PushDown( o );
 91     if ( R <= m ) return Query ( L , R , lson ) ;
 92     if ( m <  L ) return Query ( L , R , rson ) ;
 93
 94     return Query ( L , R , lson ) | Query ( L , R , rson ) ;
 95 }
 96
 97
 98 void Orz()
 99 {
100     int N,M;
101     int l, r, co;
102     char ch;
103     while( ~scanf("%d%d",&N,&M) && (N+M) )
104     {
105         getchar();
106         CLR( col, 0 ), CLR( lazy, 0 );
107         Build(1,N,1);
108         //Update(1,N,1,1,N,1);
109         REP( i,0,M )
110         {
111             ch = getchar();
112             if( ch == ‘P‘ )
113             {
114                 scanf( "%d %d %d", &l, &r, &co );
115                 Update( l, r, co-1, 1, N, 1 );
116             }
117             else
118             {
119                 scanf( "%d %d", &l, &r );
120                 int ans = Query(l,r,1,N,1);
121                 int flag = 0;
122                 REP( i, 0, 30 )
123                 {
124                     if( ans & (1 << i) )
125                     {
126                         if( flag )    printf(" ");
127                         flag = 1;
128                         printf( "%d", i + 1 );
129                     }
130                 }
131                 puts("");
132             }
133             getchar();
134         }
135     }
136 }
137
138
139 int main()
140 {
141     #ifdef  BUG
142         freopen( "in.txt", "r", stdin );
143     #endif
144     Orz();
145     return 0;
146 }

代码君

分析②: col[o] 存放当前线段内的颜色，

HDU 5023 A Corrupt Mayor's Performance Art