HDOJ2665 &&poj2104 k-th number（主席树+求第K小）

转自：http://www.cnblogs.com/Empress/p/4652449.html

题意：给n、m 

　　   下面有n个数 (编号1到n)

　　　有m个询问，询问的是上面的数的编号在[l,r]之间第k小的数 

 

n、m的范围都是

HDOJ2665的题意居然是一样的~~~说好的第K大呢···

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define lson l, m#define rson m+1, rconst int N = 1e5 + 5;int L[N << 5], R[N << 5], sum[N << 5]; //这个区间内数字的个数int tot; //tot是现在数组开到多大了int a[N], T[N], Hash[N];//Ti表示一棵[1, i]区间的线段树//我们并不需要建一整棵树，我们只需要 单独建立logn个结点，跟Ti−1连起来就好了int build(int l, int r){int rt = (++tot);sum[rt] = 0;if (l<r){int m = (l + r) >> 1;L[rt] = build(lson);R[rt] = build(rson);}return rt;}int update(int pre, int l, int r, int x){int rt = (++tot);L[rt] = L[pre], R[rt] = R[pre], sum[rt] = sum[pre] + 1;if (l<r){int m = (l + r) >> 1;if (x <= m)L[rt] = update(L[pre], lson, x);elseR[rt] = update(R[pre], rson, x);}return rt;}int query(int u, int v, int l, int r, int k){if (l >= r)return l;int m = (l + r) >> 1;int num = sum[L[v]] - sum[L[u]];//[u,v]=[1,v]-[1,u]if (num >= k)return query(L[u], L[v], lson, k);elsereturn query(R[u], R[v], rson, k - num);}//一边做减法 一边查询int main(){tot = 0;int n, m;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++){scanf("%d", &a[i]);Hash[i] = a[i];}sort(Hash + 1, Hash + n + 1);int d = unique(Hash + 1, Hash + n + 1) - Hash - 1;T[0] = build(1, d);for (int i = 1; i <= n; i++){int x = lower_bound(Hash + 1, Hash + d + 1, a[i]) - Hash;T[i] = update(T[i - 1], 1, d, x);}while (m--){int l, r, k;scanf("%d%d%d", &l, &r, &k);int x = query(T[l - 1], T[r], 1, d, k);printf("%d\n", Hash[x]);}return 0;}

转自：http://www.cnblogs.com/Empress/p/4652449.html