POJ 3468 A Simple Problem with Integers

题目链接：http://poj.org/problem?id=3468

题意：给N个数，有两组操作：

(1)对某个区间[a, b]中所有的数都增加c

(2)求某个区间[a, b]个数的和

共操作Q次，每次只输出操作(2)的结果

 

第二棵线段树……TUT……

需要用到延迟标记，每次更新不用到达叶子节点，只需要在标记上进行累加，当询问的时候再将标记向下传递。

 

  1 #include <cstdio>

  2 #include <cstdlib>

  3 

  4 const int MAXN = 100000 + 5;

  5 

  6 struct TNode

  7 {

  8     int l, r;

  9     long long int sum;

 10     long long int mark;

 11     TNode *l_child, *r_child;

 12 };

 13 

 14 TNode Tree[ MAXN + MAXN ];

 15 int cnt;

 16 long long int Sum;

 17 

 18 void Build( int l, int r, TNode *Td )

 19 {

 20     Td -> l = l;

 21     Td -> r = r;

 22 

 23     Td -> mark = 0;

 24 

 25     if ( l == r )

 26     {

 27         scanf("%lld", &Td->sum );

 28         return;

 29     }

 30 

 31     ++cnt;

 32     Td -> l_child = &Tree[cnt];

 33     Build( l, ( l + r ) / 2, Td -> l_child );

 34 

 35     ++cnt;

 36     Td -> r_child = &Tree[cnt];

 37     Build( ( l + r ) / 2 + 1, r, Td -> r_child );

 38 

 39     Td -> sum = Td -> l_child -> sum + Td -> r_child -> sum;

 40 

 41     return;

 42 }

 43 

 44 void Update( int st, int ed, int &e, TNode *Td )

 45 {

 46     if ( st == Td -> l && ed == Td -> r )

 47     {

 48         Td -> mark += e;

 49         return;

 50     }

 51 

 52     Td -> sum += e * ( ed - st + 1 );

 53 

 54     int mid = ( Td -> l + Td -> r ) / 2;

 55 

 56     if ( ed <= mid )

 57         Update( st, ed, e, Td -> l_child );

 58     else if ( st > mid )

 59         Update( st, ed, e, Td -> r_child );

 60     else

 61     {

 62         Update( st, mid, e, Td -> l_child );

 63         Update( mid + 1, ed, e, Td -> r_child );

 64     }

 65 }

 66 

 67 void Query( int st, int ed, TNode *Td )

 68 {

 69     if ( st == Td -> l && ed == Td -> r )

 70     {

 71         Sum += Td ->sum + Td -> mark * ( ed - st + 1 );

 72         return;

 73     }

 74 

 75     if ( Td -> mark )

 76     {

 77         Td -> l_child -> mark += Td ->mark;

 78         Td -> r_child -> mark += Td ->mark;

 79         Td -> sum += Td -> mark * ( Td -> r - Td -> l + 1 );

 80         Td -> mark = 0;

 81     }

 82 

 83     int mid = ( Td -> l + Td -> r ) / 2;

 84 

 85     if ( ed <= mid )

 86         Query( st, ed, Td->l_child );

 87     else if ( st > mid )

 88         Query( st, ed, Td->r_child );

 89     else

 90     {

 91         Query( st, mid, Td->l_child );

 92         Query( mid + 1, ed, Td->r_child );

 93     }

 94 }

 95 

 96 int main()

 97 {

 98     int N, Q;

 99     while ( scanf( "%d%d", &N, &Q ) != EOF )

100     {

101         Build( 1, N, Tree );

102 

103         char ch;

104         int a, b, e;

105         for ( int i = 0; i < Q; ++i )

106         {

107             getchar();

108             ch = getchar();

109             switch ( ch )

110             {

111             case 'Q':

112                 Sum = 0;

113                 scanf( "%d%d", &a, &b );

114                 Query( a, b, Tree );

115                 printf( "%lld\n", Sum );

116                 break;

117 

118             case 'C':

119                 scanf( "%d%d%d", &a, &b, &e );

120                 Update( a, b, e, Tree );

121                 break;

122             }

123         }

124     }

125     return 0;

126 }