K-th Number
Time Limit: 20000ms
Case Time Limit: 2000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 210464-bit integer IO format: %lld Java class name: Main
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3
Sample Output
5 6 3
Source
解题：解决这种问题其实最快的 划分树，
好吧，初学函数式线段树，水一题。。。
可以加减？其实做减法就是说在此此间有多少个数加入到子数中。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 100010; 7 struct Node { 8 int L,R,sum; 9 } tree[maxn<<5]; 10 struct A{ 11 int x,id; 12 bool operator<(const A& rhs) const{ 13 return x < rhs.x; 14 } 15 }a[maxn]; 16 int root[maxn],rk[maxn],tot,n,m; 17 void update(int val,int &v,int L,int R) { 18 tree[tot] = tree[v]; 19 v = tot++; 20 ++tree[v].sum; 21 if(L == R) return; 22 int mid = (L + R)>>1; 23 if(val <= mid) update(val,tree[v].L,L,mid); 24 else update(val,tree[v].R,mid+1,R); 25 } 26 int query(int i,int j,int k,int L,int R) { 27 if(L == R) return L; 28 int tmp = tree[tree[j].L].sum - tree[tree[i].L].sum; 29 int mid = (L + R)>>1; 30 if(k <= tmp) return query(tree[i].L,tree[j].L,k,L,mid); 31 else return query(tree[i].R,tree[j].R,k - tmp,mid+1,R); 32 } 33 int main() { 34 while(~scanf("%d%d",&n,&m)){ 35 for(int i = 1; i <= n; ++i){ 36 scanf("%d",&a[i].x); 37 a[i].id = i; 38 } 39 sort(a+1,a+1+n); 40 for(int i = 1; i <= n; ++i) 41 rk[a[i].id] = i; 42 tot = 1; 43 memset(tree,0,sizeof tree); 44 root[0] = 0; 45 for(int i = 1; i <= n; ++i){ 46 root[i] = root[i-1]; 47 update(rk[i],root[i],1,n); 48 } 49 while(m--){ 50 int i,j,k; 51 scanf("%d%d%d",&i,&j,&k); 52 printf("%d ",a[query(root[i-1],root[j],k,1,n)].x); 53 } 54 } 55 return 0; 56 }
整体二分。。。为整体二分答案，离线处理 假使我们的答案是mid 那么就把比不比mid大的元素对应的下标所在树状数组中的位置+1.数状数组查询区间 如果个数不少于k，那么么k大就假设是mid，然后我们再把mid缩小，看看是不是还满足，这样即使第一次求出来的k大不对，由于后面把这一次已经假设求出k大的放到左边，缩小mid，再次查询，所以，可以保证最后的k大是正确的
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 const int INF = 0x3f3f3f3f; 6 const int maxn = 100010; 7 struct QU { 8 int l,r,k; 9 } Q[maxn]; 10 struct DATA { 11 int val,id; 12 bool operator<(const DATA &t) const { 13 return val < t.val; 14 } 15 } d[maxn]; 16 int C[maxn],ans[maxn],id[maxn],tmp[maxn],n,m,p; 17 bool vis[maxn]; 18 void add(int i,int val) { 19 while(i < maxn) { 20 C[i] += val; 21 i += i&-i; 22 } 23 } 24 int sum(int i,int ret = 0) { 25 while(i > 0) { 26 ret += C[i]; 27 i -= i&-i; 28 } 29 return ret; 30 } 31 void solve(int lt,int rt,int L,int R) { 32 if(lt > rt || L == R) return; 33 int mid = (L + R)>>1,cnt = 0; 34 while(p < n && d[p + 1].val <= mid) add(d[++p].id,1); 35 while(d[p].val > mid) add(d[p--].id,-1); 36 for(int i = lt; i <= rt; ++i) { 37 if(sum(Q[id[i]].r) - sum(Q[id[i]].l-1) >= Q[id[i]].k) { 38 ans[id[i]] = mid; 39 vis[i] = true; 40 ++cnt; 41 } else vis[i] = 0; 42 } 43 int a = lt,b = lt + cnt; 44 for(int i = lt; i <= rt; ++i) 45 if(vis[i]) tmp[a++] = id[i]; 46 else tmp[b++] = id[i]; 47 for(int i = lt; i <= rt; ++i) id[i] = tmp[i]; 48 solve(lt,a-1,L,mid); 49 solve(a,b-1,mid+1,R); 50 } 51 int main() { 52 int low,high; 53 while(~scanf("%d%d",&n,&m)) { 54 low = INF; 55 high = -INF; 56 memset(C,0,sizeof C); 57 memset(vis,false,sizeof vis); 58 for(int i = 1; i <= n; ++i) { 59 scanf("%d",&d[i].val); 60 d[i].id = i; 61 low = min(low,d[i].val); 62 high = max(high,d[i].val); 63 } 64 for(int i = 1; i <= m; ++i) { 65 id[i] = i; 66 scanf("%d%d%d",&Q[i].l,&Q[i].r,&Q[i].k); 67 } 68 sort(d + 1,d + n + 1); 69 solve(1,m,low,high + 1); 70 for(int i = 1; i <= m; ++i) 71 printf("%d ",ans[i]); 72 } 73 return 0; 74 }
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 const int maxn = 100010; 6 const int INF = 0x3f3f3f3f; 7 struct DATA{ 8 int val,pos; 9 bool operator<(const DATA &rhs)const{ 10 return val < rhs.val; 11 } 12 }d[maxn]; 13 struct QU{ 14 int l,r,k,id; 15 }Q[maxn],A[maxn],B[maxn]; 16 int c[maxn],ans[maxn],n,m,p; 17 void add(int i,int val){ 18 while(i < maxn){ 19 c[i] += val; 20 i += i&-i; 21 } 22 } 23 int sum(int i,int ret = 0){ 24 while(i > 0){ 25 ret += c[i]; 26 i -= i&-i; 27 } 28 return ret; 29 } 30 void solve(int lt,int rt,int L,int R){ 31 if(lt > rt || L == R) return; 32 int mid = (L + R)>>1,a = 0,b = 0; 33 while(p < n && d[p+1].val <= mid) add(d[++p].pos,1); 34 while(p && d[p].val > mid) add(d[p--].pos,-1); 35 for(int i = lt; i <= rt; ++i){ 36 if(sum(Q[i].r) - sum(Q[i].l-1) >= Q[i].k){ 37 ans[Q[i].id] = mid; 38 A[a++] = Q[i]; 39 }else B[b++] = Q[i]; 40 } 41 for(int i = 0; i < a; ++i) Q[lt + i] = A[i]; 42 for(int i = 0; i < b; ++i) Q[lt + a + i] = B[i]; 43 solve(lt,lt + a - 1,L,mid); 44 solve(lt + a,rt,mid+1,R); 45 } 46 int main(){ 47 while(~scanf("%d%d",&n,&m)){ 48 int low = INF,high = -INF; 49 for(int i = 1; i <= n; ++i){ 50 scanf("%d",&d[i].val); 51 d[i].pos = i; 52 low = min(low,d[i].val); 53 high = max(high,d[i].val); 54 } 55 memset(c,0,sizeof c); 56 sort(d + 1,d + n + 1); 57 for(int i = 0; i < m; ++i){ 58 scanf("%d%d%d",&Q[i].l,&Q[i].r,&Q[i].k); 59 Q[i].id = i; 60 } 61 p = 0; 62 solve(0,m-1,low,INF); 63 for(int i = 0; i < m; ++i) 64 printf("%d ",ans[i]); 65 } 66 return 0; 67 }