HDU 4358 Boring counting 树状数组+思路

研究了整整一天orz……直接上官方题解神思路

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <vector>

  5 #include <algorithm>

  6 

  7 using namespace std;

  8 

  9 const int MAXN = 100100;

 10 

 11 struct node

 12 {

 13     int v, next;

 14 };

 15 

 16 struct subTree

 17 {

 18     int st, ed;

 19 };

 20 

 21 struct Queryy

 22 {

 23     int i;

 24     int st, ed;

 25 };

 26 

 27 int N, K, Q;

 28 int EdgeN, TimeFlag;

 29 node D[MAXN];   //树节点

 30 int C[MAXN];    //树状数组

 31 int head[MAXN];

 32 subTree tree[MAXN];   //子树映射成线性序列之后对应的区间

 33 Queryy qry[MAXN];     //查询信息

 34 int weight[MAXN], addr[MAXN];

 35 int ans[MAXN];

 36 vector<int> pos[MAXN];

 37 

 38 void AddEdge( int u, int v )

 39 {

 40     D[EdgeN].v = v;

 41     D[EdgeN].next = head[u];

 42     head[u] = EdgeN++;

 43     return;

 44 }

 45 

 46 void DFS( int cur )                    //树形结构转线性结构

 47 {

 48     tree[ cur ].st = ++TimeFlag;

 49     addr[ TimeFlag ] = weight[ cur ];

 50     for ( int i = head[cur]; i != -1; i = D[i].next )

 51         DFS( D[i].v );

 52     tree[cur].ed = TimeFlag;

 53     return;

 54 }

 55 

 56 int lowbit( int x )

 57 {

 58     return (-x) & x;

 59 }

 60 

 61 void add( int x, int val )

 62 {

 63     while ( x <= N )

 64     {

 65         C[x] += val;

 66         x += lowbit(x);

 67     }

 68     return;

 69 }

 70 

 71 int query( int x )

 72 {

 73     int res = 0;

 74     while ( x > 0 )

 75     {

 76         res += C[x];

 77         x -= lowbit(x);

 78     }

 79     return res;

 80 }

 81 

 82 bool cmp( int a, int b )

 83 {

 84     return weight[a] < weight[b];

 85 }

 86 

 87 void init()    //将weight离散化

 88 {

 89     sort( addr + 1, addr + N + 1, cmp );

 90 

 91     int cnt = 0, pre = -1;

 92     for ( int i = 1; i <= N; ++i )

 93     {

 94         if ( weight[ addr[i] ] != pre )

 95             pre = weight[ addr[i] ], weight[ addr[i] ] = ++cnt;

 96         else weight[ addr[i] ] = cnt;

 97     }

 98     return;

 99 }

100 

101 bool cmp2( Queryy a, Queryy b )

102 {

103     return a.ed < b.ed;

104 }

105 

106 void solved()

107 {

108     for ( int i = 0; i <= N; ++i ) pos[i].clear();

109 

110     sort( qry, qry + Q, cmp2 );

111     int cur = 0;

112     for ( int i = 1; i <= N; ++i )

113     {

114         int val = addr[i];

115         pos[val].push_back(i);

116         int sz = pos[val].size();

117         if ( sz == K )

118             add( pos[val][ sz - K ], 1 );

119         else if ( sz > K )

120         {

121             add( pos[val][ sz - K ], 1 );

122             add( pos[val][ sz - K - 1 ], -2 );

123         }

124         //printf( "ed = %d\n", qry[cur].ed );

125         while ( cur < Q && qry[cur].ed == i )

126         {

127             int id = qry[cur].i;

128             ans[id] = query( qry[cur].ed ) - query( qry[cur].st - 1 );

129           //  printf("ans[%d] = %d\n", id, ans[id] );

130             ++cur;

131         }

132     }

133     return;

134 }

135 

136 int main()

137 {

138     int T, cas = 0;

139     scanf( "%d", &T );

140     while ( T-- )

141     {

142         memset( head, -1, sizeof( head ) );

143         memset( C, 0, sizeof(C) );

144 

145         scanf( "%d%d", &N, &K );

146         for ( int i = 1; i <= N; ++i )

147         {

148             scanf( "%d", &weight[i] );

149             addr[i] = i;

150         }

151         init();

152 

153         EdgeN = 0;

154 

155         for ( int i = 1; i < N; ++i )      //建树

156         {

157             int u, v;

158             scanf( "%d%d", &u, &v );

159             AddEdge( u, v );

160         }

161 

162         TimeFlag = 0;

163         DFS(1);                           //树形结构转线性结构

164 

165         scanf( "%d", &Q );

166         printf( "Case #%d:\n", ++cas );

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

168         {

169             int u;

170             scanf( "%d", &u );

171             qry[i].i = i;

172             qry[i].st = tree[u].st;

173             qry[i].ed = tree[u].ed;

174            // printf( "%d %d\n", qry[i].st, qry[i].ed );

175         }

176         solved();

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

178             printf( "%d\n", ans[i] );

179 

180         if ( T ) puts("");

181     }

182     return 0;

183 }