POJ 1442 Black Box treap求區間第k大
來源：程序員人生 發布時間：2014-09-09 05:20:08 閱讀次數：3449次
題目來源：POJ 1442 Black Box
題意：輸入xi 輸出前xi個數的第i大的數
思路：試了下自己的treap模版
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
struct Node
{
Node *ch[2];
int r;
int v;
int s;
Node(){}
Node(int v): v(v) {
ch[0] = ch[1] = NULL; r = rand(); s = 1;
}
bool operator < (const Node& rhs) const{
return r < rhs.r;
}
int cmp(int x) const{
if(x == v) return -1;
return x < v ? 0 : 1;
}
void maintain(){
s = 1;
if(ch[0] != NULL) s += ch[0]->s;
if(ch[1] != NULL) s += ch[1]->s;
}
};
void rotate(Node* &o, int d){
Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o;
o->maintain(); k->maintain(); o = k;
}
void insert(Node* &o, int x){
if(o == NULL){
o = new Node(x);
}
else{
int d = (x < o->v ? 0 : 1);
insert(o->ch[d], x);
if((o->ch[d]->r) > (o->r)) rotate(o, d^1);
}
o->maintain();
}
void remove(Node* &o, int x){
int d= o->cmp(x);
if(d == -1){
Node* u = o;
if(o->ch[0] != NULL && o->ch[1] != NULL){
int d2 = (o->ch[0] > o->ch[1] ? 1 : 0);
rotate(o, d2); remove(o->ch[d2], x);
}
else{
if(o->ch[0] == NULL) o = o->ch[1];
else o = o->ch[0];
delete u;
}
}
else
remove(o->ch[d], x);
if(o != NULL) o->maintain();
}
int kth(Node* o, int k){
if(o == NULL || k <= 0 || k > o->s)
return 0;
int s = (o->ch[0] == NULL ? 0 : o->ch[0]->s);
if(k == s+1) return o->v;
else if(k <= s) return kth(o->ch[0], k);
else return kth(o->ch[1], k-s-1);
}
void removetree(Node* &x){
if(x->ch[0] != NULL) removetree(x->ch[0]);
if(x->ch[1] != NULL) removetree(x->ch[1]);
delete x;
x = NULL;
}
int n, m, a[30010];
Node *rt = NULL;
int main()
{
while(scanf("%d %d", &n, &m) != EOF)
{
srand(time(0));
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int l = 1;
for(int i = 1; i <= m; i++)
{
int x;
scanf("%d", &x);
while(l <= x)
{
insert(rt, a[l]);
l++;
}
printf("%d
", kth(rt, i));
}
//removetree(rt);
}
return 0;
}
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