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 }
