| Description Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other. Your task is to help Jiajia calculate which dog ate the food after each feeding.Input The first line contains n and m, indicates the number of dogs and the number of feedings. The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier. Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding. You can assume that n<100001 and m<50001.Output Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding. Sample Input 7 21 5 2 6 3 7 41 5 32 7 1 Sample Output 32 |
求区间第k大的值，因为题目要求了不会出现重复喂某只狗的情况，所以不需要动态更新数组，直接纯的划分树模板。
/** 划分树（查询区间第k大）*/#include#include #include using namespace std;const int MAXN = 100010;int tree[20][MAXN];//表示每层每个位置的值int sorted[MAXN];//已经排序好的数int toleft[20][MAXN];//toleft[p][i]表示第i层从1到i有数分入左边void build(int l,int r,int dep){ if(l == r)return; int mid = (l+r)>>1; int same = mid - l + 1;//表示等于中间值而且被分入左边的个数 for(int i = l; i <= r; i++) //注意是l,不是one if(tree[dep][i] < sorted[mid]) same--; int lpos = l; int rpos = mid+1; for(int i = l; i <= r; i++) { if(tree[dep][i] < sorted[mid]) tree[dep+1][lpos++] = tree[dep][i]; else if(tree[dep][i] == sorted[mid] && same > 0) { tree[dep+1][lpos++] = tree[dep][i]; same--; } else tree[dep+1][rpos++] = tree[dep][i]; toleft[dep][i] = toleft[dep][l-1] + lpos - l; } build(l,mid,dep+1); build(mid+1,r,dep+1);}//查询区间第k大的数,[L,R]是大区间，[l,r]是要查询的小区间int query(int L,int R,int l,int r,int dep,int k){ if(l == r)return tree[dep][l]; int mid = (L+R)>>1; int cnt = toleft[dep][r] - toleft[dep][l-1]; if(cnt >= k) { int newl = L + toleft[dep][l-1] - toleft[dep][L-1]; int newr = newl + cnt - 1; return query(L,mid,newl,newr,dep+1,k); } else { int newr = r + toleft[dep][R] - toleft[dep][r]; int newl = newr - (r-l-cnt); return query(mid+1,R,newl,newr,dep+1,k-cnt); }}int main(){ int n,m; while(scanf("%d%d",&n,&m)==2) { memset(tree,0,sizeof(tree)); for(int i = 1; i <= n; i++) { scanf("%d",&tree[0][i]); sorted[i] = tree[0][i]; } sort(sorted+1,sorted+n+1); build(1,n,0); int s,t,k; while(m--) { scanf("%d%d%d",&s,&t,&k); printf("%d\n",query(1,n,s,t,0,k)); } } return 0;}