学步园

from : http://www.notonlysuccess.com/?p=142

poj 2104 代码如下：

#include <cstdio><br /> #include <algorithm><br /> using namespace std;<br /> #define M 100001<br /> #define LL(x) ((x)<<1)<br /> #define RR(x) ((x)<<1|1)<br /> struct Seg_Tree{<br /> int left,right;<br /> int mid() {<br /> return (left + right) >> 1;<br /> }<br /> }tt[M*4];<br /> int len;<br /> int sorted[M];<br /> int toLeft[20][M];<br /> int val[20][M];<br /> void build(int l, int r, int d, int idx) {<br /> tt[idx].left = l;<br /> tt[idx].right = r;<br /> if(tt[idx].left == tt[idx].right)<br /> return ;<br /> int mid = tt[idx].mid();<br /> int lsame = mid - l + 1;//lsame表示和val_mid相等且分到左边的<br /> for(int i = l ; i <= r ; i ++) {<br /> if(val[d][i] < sorted[mid]) {<br /> 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<br /> }<br /> }<br /> int lpos = l;<br /> int rpos = mid+1;<br /> int same = 0;<br /> for(int i = l; i <= r; i ++) {<br /> if(i == l) {<br /> toLeft[d][i] = 0;//toLeft[i]表示[ tt[idx].left , i ]区域里有多少个数分到左边<br /> } else {<br /> toLeft[d][i] = toLeft[d][i-1];<br /> }<br /> if(val[d][i] < sorted[mid]) {<br /> toLeft[d][i] ++;<br /> val[d+1][lpos++] = val[d][i];<br /> } else if(val[d][i] > sorted[mid]) {<br /> val[d+1][rpos++] = val[d][i];<br /> } else {<br /> if(same < lsame) {//有lsame的数是分到左边的<br /> same ++;<br /> toLeft[d][i] ++;<br /> val[d+1][lpos++] = val[d][i];<br /> } else {<br /> val[d+1][rpos++] = val[d][i];<br /> }<br /> }<br /> }<br /> build(l,mid,d+1,LL(idx));<br /> build(mid+1,r,d+1,RR(idx));<br /> }<br /> int query(int l,int r,int k,int d,int idx) {<br /> if(l == r) {<br /> return val[d][l];<br /> }<br /> int s;//s表示[ l , r ]有多少个分到左边<br /> int ss;//ss表示 [tt[idx].left , l-1 ]有多少个分到左边<br /> if(l == tt[idx].left) {<br /> s = toLeft[d][r];<br /> ss = 0;<br /> } else {<br /> s = toLeft[d][r] - toLeft[d][l-1];<br /> ss = toLeft[d][l-1];<br /> }<br /> if(s >= k) {//有多于k个分到左边,显然去左儿子区间找第k个<br /> int newl = tt[idx].left + ss;<br /> int newr = tt[idx].left + ss + s - 1;//计算出新的映射区间<br /> return query(newl,newr,k,d+1,LL(idx));<br /> } else {<br /> int mid = tt[idx].mid();<br /> int bb = l - tt[idx].left - ss;//bb表示 [tt[idx].left , l-1 ]有多少个分到右边<br /> int b = r - l + 1 - s;//b表示 [l , r]有多少个分到右边<br /> int newl = mid + bb + 1;<br /> int newr = mid + bb + b;<br /> return query(newl,newr,k-s,d+1,RR(idx));<br /> }<br /> }<br /> int main()<br /> {<br /> int n, m;<br /> scanf("%d%d",&n,&m);<br /> for(int i = 1; i < n+1; i++)<br /> {<br /> scanf("%d",&val[0][i]);<br /> sorted[i] = val[0][i];<br /> }<br /> sort(sorted + 1 , sorted + n + 1);<br /> build(1,n,0,1);<br /> while(m --)<br /> {<br /> int l,r,k;<br /> scanf("%d%d%d", &l, &r, &k);<br /> printf("%d/n", query(l,r,k,0,1));<br /> }<br /> return 0;<br /> }<br />