划分树-POJ2104,POJ2761,HDU2665,HDU3743

　　做了几道划分树的题目，划分树的效率很高，用来求区间第K小值和区间小于K小值的和，每次动态询问的复杂度为O(logn)。

其中POJ2104,POJ2761,HDU2665为典型的求区间第K小值，HDU3743还加上了求和的操作，推荐去做一做。

　　下面是HUD3743的代码：

  1 //STATUS:C++_AC_343MS_30048KB

  2 #include<stdio.h>

  3 #include<stdlib.h>

  4 #include<string.h>

  5 #include<math.h>

  6 #include<iostream>

  7 #include<string>

  8 #include<algorithm>

  9 #include<vector>

 10 #include<queue>

 11 #include<stack>

 12 #include<map>

 13 using namespace std;

 14 #define LL __int64

 15 #define pii pair<int,int>

 16 #define mem(a,b) memset(a,b,sizeof(a))

 17 #define lson l,mid,rt<<1

 18 #define rson mid+1,r,rt<<1|1

 19 #define PI acos(-1.0)

 20 const int N=100010,INF=0x3f3f3f3f,MOD=10000,STA=8000010;

 21 //const LL LNF=0x3f3f3f3f3f3f3f3f;

 22 const double DNF=1e13;

 23 //

 24 inline int Max(int a,int b){return a>b?a:b;}

 25 inline int Min(int a,int b){return a<b?a:b;}

 26 void swap(int& a,int& b){int t=a;a=b;b=t;}

 27 void swap(LL& a,LL& b){LL t=a;a=b;b=t;}

 28 //

 29 

 30 int val[N],num[20][N],cnt[20][N];

 31 LL sum[20][N],s[N];

 32 int T,n,m,knum;

 33 LL ksum;

 34 

 35 void build(int u,int l,int r)

 36 {

 37     if(l==r){

 38         sum[u][l]=num[u][l];

 39         return;

 40     }

 41     int i,mid,midnum,kl,kr,lsame;

 42     LL s=0;

 43     mid=(l+r)>>1;

 44     kl=l;kr=mid+1;lsame=mid-l+1;

 45     midnum=val[mid];

 46     for(i=l;i<=mid;i++)   //注意这里需要统计等于中位数放在左儿子区间的个数

 47         if(val[i]<midnum)lsame--;

 48     for(i=l;i<=r;i++){

 49         if(num[u][i]<midnum || (num[u][i]==midnum && lsame)){  //注意等于中位数情况

 50             if(num[u][i]==midnum)lsame--;

 51             num[u+1][kl++]=num[u][i];

 52             sum[u][i]=s+(LL)num[u][i];

 53             s=sum[u][i];

 54         }

 55         else {

 56             num[u+1][kr++]=num[u][i];

 57             sum[u][i]=s;

 58         }

 59         cnt[u][i]=kl-l;

 60     }

 61     build(u+1,l,mid);

 62     build(u+1,mid+1,r);

 63 }

 64 

 65 void query(int u,int l,int r,int a,int b,int k)

 66 {

 67     if(a==b){   //注意这里可能l<r啦

 68         knum=num[u][a];   

 69         ksum+=num[u][a];

 70         return;

 71     }

 72     int i,t,mid,cnta;

 73     mid=(l+r)>>1;

 74     cnta=(a>l?cnt[u][a-1]:0);   //注意l==a

 75     t=cnt[u][b]-cnta;

 76     if(k<=t){

 77         query(u+1,l,mid,l+cnta,l+cnt[u][b]-1,k);

 78     }

 79     else{

 80         ksum+=sum[u][b]-(a>l?sum[u][a-1]:0);

 81         query(u+1,mid+1,r,mid+a-l-cnta+1,mid+b-l-cnt[u][b]+1,k-t);

 82     }

 83 }

 84 

 85 int main()

 86 {

 87   //  freopen("in.txt","r",stdin);

 88     int i,j,sz=1,a,b,k;

 89     LL ans;

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

 91     while(T--)

 92     {

 93         mem(num,0);

 94         scanf("%d",&n);

 95         s[0]=0;

 96         for(i=1;i<=n;i++){

 97             scanf("%d",&val[i]);

 98             num[0][i]=val[i];

 99             s[i]=s[i-1]+val[i];

100         }

101         sort(val+1,val+n+1);

102         build(0,1,n);

103 

104         scanf("%d",&m);

105         printf("Case #%d:\n",sz++);

106         while(m--){

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

108             a++,b++;

109             k=(b-a)/2+1;

110             ksum=0;

111             query(0,1,n,a,b,k);

112             ans=(LL)knum*k-ksum;

113             ans+=s[b]-s[a-1]-ksum-(LL)knum*(b-a+1-k);

114             printf("%I64d\n",ans);

115         }

116         putchar('\n');

117     }

118     return 0;

119 }