题目:http://poj.org/problem?id=1442
简单的插入、查询第k大数的题。这本来应该算一道经典的堆的应用题,结果我用SBT水过去了。。。算了,有机会要用堆实现一下。
SBT 版:
顺便说下,用cin一次TLE一次1000MS惊现过去,换成scanf 400MS无压力。。。(跟其他人比还是太慢了。。。)
POJ 1442 SBT
#include <fstream>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <iomanip>
#include <climits>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;
const int maxn= 100010;
class SBT{
public:
void Clear(){
memset(K,0,sizeof(K));
memset(L,0,sizeof(L));
memset(R,0,sizeof(R));
memset(S,0,sizeof(S));
RT= SZ= 0;
}
void Insert(int key) {Insert(RT,key);}
int Delete(int key) {return Delete(RT,key);}
int Succ(int key) {return Succ(RT,key);}
int Pred(int key) {return Pred(RT,key);}
int Rank(int key) {return Rank(RT,key);}
int Search(int key) {return Search(RT,key);}
int Select(int key) {return Select(RT,key);}
private:
int K[maxn];
int L[maxn];
int R[maxn];
int S[maxn];
int RT, SZ;
void LeftRotate(int &x){
int k= R[x];
R[x]= L[k];
L[k]= x;
S[k]= S[x];
S[x]= S[L[x]]+S[R[x]]+1;
x= k;
}
void RightRotate(int &x){
int k= L[x];
L[x]= R[k];
R[k]= x;
S[k]= S[x];
S[x]= S[L[x]]+S[R[x]]+1;
x= k;
}
void MaintainFat(int &t){
if (S[L[L[t]]]>S[R[t]]){
RightRotate(t);
MaintainFat(R[t]);
MaintainFat(t);
return;
}
if (S[R[L[t]]]>S[R[t]]){
LeftRotate(L[t]);
RightRotate(t);
MaintainFat(L[t]);
MaintainFat(R[t]);
MaintainFat(t);
return;
}
if (S[R[R[t]]]>S[L[t]]){
LeftRotate(t);
MaintainFat(L[t]);
MaintainFat(t);
return;
}
if (S[L[R[t]]]>S[L[t]]){
RightRotate(R[t]);
LeftRotate(t);
MaintainFat(L[t]);
MaintainFat(R[t]);
MaintainFat(t);
return;
}
}
void Maintain(int &t, int flag){
if (!flag){
if (S[L[L[t]]] >S[R[t]])
RightRotate(t);
else if (S[R[L[t]]] >S[R[t]]){
LeftRotate(L[t]);
RightRotate(t);
}
else
return;
}
else{
if (S[R[R[t]]] >S[L[t]])
LeftRotate(t);
else if (S[L[R[t]]] >S[L[t]]){
RightRotate(R[t]);
LeftRotate(t);
}
else
return;
}
Maintain(L[t], false);
Maintain(R[t], true);
Maintain(t, true);
Maintain(t, false);
}
void Insert(int &t, int key){
if (t==0){
t= ++SZ;
K[t]= key;
S[t]= 1;
return;
}
S[t]++;
if (key<K[t])
Insert(L[t],key);
else
Insert(R[t],key);
Maintain(t,key>K[t]);
}
int Delete(int &t, int key){
S[t]--;
if ((key==K[t])||(key<K[t]&&L[t]==0)||(key>K[t]&&R[t]==0)){
int ret= K[t];
if (L[t]==0 || R[t]==0)
t= L[t]+R[t]; // T change to his Leftson or Rightson
else
K[t]= Delete(L[t], K[t]+1); // Not find then delete the last find Point
return ret;
}
else{
if (key<K[t])
return Delete(L[t],key);
else
return Delete(R[t],key);
}
}
int Search(int t, int key){ //return root point
if (t==0 || key==K[t])
return t;
if (key<K[t])
return Search(L[t],key);
else
return Search(R[t],key);
}
int Select(int t, int k){ // return K-th int tree
int num= S[L[t]]+1;
if (k==num)
return K[t];
else if (k<num)
return Select(L[t],k);
else
return Select(R[t],k-num);
}
int Succ(int t, int key){
if (t==0)
return key;
if (key>=K[t]) //
return Succ(R[t], key);
else{
int r= Succ(L[t], key);
if (r==key)
return K[t];
else
return r;
}
}
int Pred(int t, int key){
if (t==0)
return key;
if (key<=K[t]) //
return Pred(L[t], key);
else {
int r= Pred(R[t], key);
if (r==key)
return K[t];
else
return r;
}
}
int Rank(int t, int key){
if (t==0)
return 1;
if (key<=K[t]) //
return Rank(L[t], key);
else if (key>K[t])
return S[L[t]]+1+Rank(R[t],key);
}
};
int main()
{
int a[30005];
int com[30005];
memset(com,0,sizeof(com));
SBT sbt;
int sbtnum=0;
int n,m;
scanf("%d%d",&m,&n);
for (int i=0;i<m;i++)
scanf("%d",&a[i]);
for (int i=0;i<n;i++)
{
int k;
scanf("%d",&k);
com[k]++;
}
int k=0;
for (int i=0;i<m;i++)
{
sbt.Insert(a[i]);
sbtnum++;
while(com[sbtnum])
{
com[sbtnum]--;
k++;
printf("%d\n",sbt.Select(k));
}
}
return 0;
}
STL priority_queue堆版:
先占个位^_^~~~
线段树版:
先占个位^_^~~~