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POJ 1442 Black Box Treap 模板题

2016-09-22 18:06 465 查看
题目链接:http://poj.org/problem?id=1442

给两个序列A,B

求A中前B[i]个数第i小的数是几

poj不支持srand(time(NULL)) RE的可能是这个原因

代码:

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#define sf scanf
#define pf printf

using namespace std;
const int maxn = 444444 + 50;

/** Treap */
int ch[maxn][2],size[maxn],fa[maxn],fix[maxn],key[maxn];
int root,sroot,tot;
inline void Treap_Init(){
sroot = root = 0;
fa[root] = sroot;
ch[root][0] = ch[root][1] = 0;
size[sroot] = 0;
tot = 1;
}

inline void PushUp(int rt){
size[rt] = 1 + size[ch[rt][0]] + size[ch[rt][1]];
}

inline int NewNode(int FA,int k){
fa[tot] = FA;
ch[tot][0] = ch[tot][1] = sroot;
size[tot] = 1;
fix[tot] = rand();
key[tot] = k;
return tot++;
}

inline void rotate(int rt,int kind){
int prt = fa[rt];
ch[prt][kind] = ch[rt][!kind];
fa[ch[rt][!kind]] = prt;

if(fa[prt]!=sroot)
ch[ fa[prt] ][ ch[fa[prt]][1] == prt ] = rt;
fa[rt] = fa[prt];

ch[rt][!kind] = prt;
fa[prt] = rt;
}

inline void Insert(int k){
int rt = root;
while( ch[rt][ (k >= key[rt]) ] != sroot){
rt = ch[rt][ (k >= key[rt]) ];
}
ch[rt][ k >= key[rt] ] = NewNode(rt,k);
rt = ch[rt][ k >= key[rt] ];
while( (fa[rt] != sroot) && (fix[rt] >= fix[fa[rt]]) ){
int prt = fa[rt];
rotate(rt,ch[fa[rt]][1] == rt);
PushUp(prt);
PushUp(rt);
}
if(fa[rt] == sroot) root = rt;
while( fa[rt] != sroot ){
rt = fa[rt];
PushUp(rt);
}
}

inline int Search(int k){
int rt = root;
while( true ){
if(k <= size[ch[rt][0]]) rt = ch[rt][0];
else if(k > size[ch[rt][0]] + 1) {
k -= size[ch[rt][0]] + 1;
rt = ch[rt][1];
}
else return key[rt];
}
}

int num[maxn];
int main(){
//    freopen("in.txt","r",stdin);
int n,m,cur,tmp;
while( ~sf("%d%d",&n,&m) ){
Treap_Init();
for(int i = 1;i <= n;++i) sf("%d",&num[i]);
cur = 0;
for(int i = 1;i <= m;++i){
sf("%d",&tmp);
while(cur < tmp){
Insert(num[++cur]);
}
pf("%d\n",Search(i));
}
}
}
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