HDU5126---stars （CDQ套CDQ套 树状数组）

题意：Q次操作，三维空间内 每个星星对应一个坐标，查询以（x1,y1,z1） (x2,y2,z2)为左下顶点 、右上顶点的立方体内的星星的个数。

注意Q的范围为50000，显然离散化之后用三维BIT会MLE。 我们可以用一次CDQ把三维变成二维，变成二维之后就有很多做法了，树套树，不会树套树的话还可以继续CDQ由二维变成一维，，变成一维了就好做了，，最基本的数据结构题目了。。

不得不说、CDQ真的很神奇。

下面做法就是CDQ套CDQ套树状数组。

  1 #include <cstdio>

  2 #include <cstdlib>

  3 #include <cstring>

  4 #include <algorithm>

  5 using namespace std;

  6 const int maxn = 50010;

  7 inline int lowbit (int x)

  8 {

  9     return x & -x;

 10 }

 11 int c[maxn*9],MAX;

 12 void add(int x,int d)

 13 {

 14     while (x <= MAX)

 15     {

 16         c[x] += d;

 17         x += lowbit(x);

 18     }

 19 }

 20 int sum(int x)

 21 {

 22     int ans = 0;

 23     while (x)

 24     {

 25         ans += c[x];

 26         x -= lowbit (x);

 27     }

 28     return ans;

 29 }

 30 struct Point

 31 {

 32     int x,y,z;

 33     int kind,idx,delt;

 34     Point() {}

 35     Point(int _x,int _y,int _z,int _delt,int _kind,int _idx):

 36         x(_x), y(_y), z(_z), delt(_delt), kind(_kind), idx(_idx) {}

 37 

 38 } star[maxn << 4],star3[maxn << 4];

 39 int ans[maxn<<1];

 40 bool cmp1(const Point &p1,const Point &p2)

 41 {

 42     return p1.x < p2.x || ((p1.x == p2.x) && (p1.idx < p2.idx) );

 43 }

 44 bool cmp2(const Point &p1,const Point &p2)

 45 {

 46     return p1.y < p2.y || ((p1.y == p2.y) && (p1.idx < p2.idx) );

 47 }

 48 void CDQ2(int l,int r)

 49 {

 50     if (l >= r)

 51         return;

 52     int mid = (l + r) >> 1;

 53     CDQ2(l,mid);

 54     CDQ2(mid+1,r);

 55     int j = l;

 56     for (int i = mid + 1; i <= r; i++)

 57     {

 58         if (star3[i].kind == 1)

 59         {

 60             for ( ; j <= mid  && (star3[j].y <= star3[i].y); j++)

 61             {

 62                 if (star3[j].kind == 0)

 63                     add(star3[j].z,star3[j].delt);

 64             }

 65             ans[star3[i].idx] += sum(star3[i].z) * star3[i].delt;

 66         }

 67     }

 68     for (int i = l; i < j; i++)

 69     {

 70         if (star3[i].kind == 0)

 71             add(star3[i].z,-star3[i].delt);

 72     }

 73     inplace_merge(star3+l,star3+mid+1,star3+r+1,cmp2);

 74 }

 75 void CDQ1(int l,int r)

 76 {

 77     if (l == r)

 78         return;

 79     int mid = (l + r) >> 1;

 80     CDQ1(l, mid);

 81     CDQ1(mid+1, r);

 82     int tot = 1;

 83     for (int j = l; j <= mid ; j++)

 84         if (star[j].kind == 0)

 85             star3[tot++] = star[j];

 86     for (int i = mid + 1; i <= r; i++)

 87     {

 88         if (star[i].kind == 1)

 89             star3[tot++] = star[i];

 90     }

 91     sort(star3+1,star3+tot,cmp1);

 92     CDQ2(1,tot-1);

 93 }

 94 int vec[maxn << 4],idx;

 95 void hash_(int tot)

 96 {

 97     sort(vec,vec+idx);

 98     idx = unique(vec,vec+idx) - vec;

 99     MAX = idx + 1;

100     for (int i = 1; i <= tot; i++)

101         star[i].z = lower_bound(vec,vec+idx,star[i].z) - vec + 1;

102 }

103 int main(void)

104 {

105 #ifndef ONLINE_JUDGE

106     freopen("in.txt","r",stdin);

107 #endif // ONLINE_JUDGE

108     int T,Q;

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

110     while (T--)

111     {

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

113         int tot = 1;

114         int totq = 0;

115         idx = 0;

116         memset(c,0,sizeof (c));

117         memset(ans,0,sizeof(ans));

118         for (int i = 1; i <= Q; i++)

119         {

120             int op,x1,y1,z1,x2,y2,z2;

121             scanf ("%d",&op);

122             if (op == 1)

123             {

124                 scanf ("%d%d%d",&x1,&y1,&z1);

125                 star[tot] = Point(x1,y1,z1,1,0,totq);

126                 vec[idx++] = z1;

127                 ans[totq] = -1;

128                 tot++;

129                 totq++;

130             }

131             if (op == 2)

132             {

133                 scanf ("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);

134                 star[tot] = Point(x1-1, y1-1, z1-1, -1, 1, totq),  vec[idx++] = z1-1, tot++;

135                 star[tot] = Point(x2,   y1-1, z1-1,  1, 1, totq),  vec[idx++] = z1-1, tot++;

136                 star[tot] = Point(x2 ,  y2  , z1-1, -1, 1, totq),  vec[idx++] = z1-1, tot++;

137                 star[tot] = Point(x1-1, y2,   z1-1,  1, 1, totq),  vec[idx++] = z1-1, tot++;

138                 star[tot] = Point(x1-1, y2,   z2  , -1, 1, totq),  vec[idx++] = z2  , tot++;

139                 star[tot] = Point(x2  , y2,   z2  ,  1, 1, totq),  vec[idx++] = z2  , tot++;

140                 star[tot] = Point(x2  , y1-1, z2  , -1, 1, totq),  vec[idx++] = z2  , tot++;

141                 star[tot] = Point(x1-1, y1-1, z2  ,  1, 1, totq),  vec[idx++] = z2  , tot++;

142                 totq++;

143             }

144         }

145         hash_(tot);

146         CDQ1(1,tot-1);

147         for (int i = 0; i < totq; i++)

148             if (~ans[i])

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

150     }

151     return 0;

152 }