第k元素 划分树
from : http://www.notonlysuccess.com/?p=142
poj 2104 代码如下：
#include <cstdio> #include <algorithm> using namespace std; #define M 100001 #define LL(x) ((x)<<1) #define RR(x) ((x)<<1|1) struct Seg_Tree{ int left,right; int mid() { return (left + right) >> 1; } }tt[M*4]; int len; int sorted[M]; int toLeft[20][M]; int val[20][M]; void build(int l, int r, int d, int idx) { tt[idx].left = l; tt[idx].right = r; if(tt[idx].left == tt[idx].right) return ; int mid = tt[idx].mid(); int lsame = mid - l + 1;//lsame表示和val_mid相等且分到左边的 for(int i = l ; i <= r ; i ++) { if(val[d][i] < sorted[mid]) { lsame --;//先假设左边的数(mid - l + 1)个都等于val_mid,然后把实际上小于val_mid的减去 } } int lpos = l; int rpos = mid+1; int same = 0; for(int i = l; i <= r; i ++) { if(i == l) { toLeft[d][i] = 0;//toLeft[i]表示[ tt[idx].left , i ]区域里有多少个数分到左边 } else { toLeft[d][i] = toLeft[d][i-1]; } if(val[d][i] < sorted[mid]) { toLeft[d][i] ++; val[d+1][lpos++] = val[d][i]; } else if(val[d][i] > sorted[mid]) { val[d+1][rpos++] = val[d][i]; } else { if(same < lsame) {//有lsame的数是分到左边的 same ++; toLeft[d][i] ++; val[d+1][lpos++] = val[d][i]; } else { val[d+1][rpos++] = val[d][i]; } } } build(l,mid,d+1,LL(idx)); build(mid+1,r,d+1,RR(idx)); } int query(int l,int r,int k,int d,int idx) { if(l == r) { return val[d][l]; } int s;//s表示[ l , r ]有多少个分到左边 int ss;//ss表示 [tt[idx].left , l-1 ]有多少个分到左边 if(l == tt[idx].left) { s = toLeft[d][r]; ss = 0; } else { s = toLeft[d][r] - toLeft[d][l-1]; ss = toLeft[d][l-1]; } if(s >= k) {//有多于k个分到左边,显然去左儿子区间找第k个 int newl = tt[idx].left + ss; int newr = tt[idx].left + ss + s - 1;//计算出新的映射区间 return query(newl,newr,k,d+1,LL(idx)); } else { int mid = tt[idx].mid(); int bb = l - tt[idx].left - ss;//bb表示 [tt[idx].left , l-1 ]有多少个分到右边 int b = r - l + 1 - s;//b表示 [l , r]有多少个分到右边 int newl = mid + bb + 1; int newr = mid + bb + b; return query(newl,newr,k-s,d+1,RR(idx)); } } int main() { int n, m; scanf("%d%d",&n,&m); for(int i = 1; i < n+1; i++) { scanf("%d",&val[0][i]); sorted[i] = val[0][i]; } sort(sorted + 1 , sorted + n + 1); build(1,n,0,1); while(m --) { int l,r,k; scanf("%d%d%d", &l, &r, &k); printf("%d/n", query(l,r,k,0,1)); } return 0; }