pat1057 stack

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了pat1057 stack相关的知识,希望对你有一定的参考价值。

超时算法,利用2的特殊性,用2个multiset来维护。单个multiset维护没法立即找到中位数。

其实也可以只用1个multiset,用一个中位指针,++,--来维护中位数。

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<map>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 7;
int n;
multiset<int> up, lo;
stack<int> sta;
void adapt() {
    int cnt = up.size() + lo.size();
    int upsz = ceil(cnt / 2.0);
    while (up.size() < upsz) {
        up.insert(*lo.begin());
        lo.erase(lo.begin());
    }
    while (up.size() > upsz) {
        int x = *up.rbegin();
        lo.insert(x);
        up.erase(up.find(x));
    }
}
void push(int x) {
    up.insert(x);
    adapt();
}
int peek() {
    return *(up.rbegin());
}
void pop(int x) {
    if (up.find(x) != up.end())
        up.erase(up.find(x));
    else
        lo.erase(lo.find(x));
    adapt();
}
int main() {
    freopen("in.txt", "r", stdin);
    cin >> n;
    while (sta.empty() == false)
        sta.pop();
    up.clear(), lo.clear();
    char cmd[13];
    int x;
    while (n--) {
        cin >> cmd;
        if (cmd[1] != ‘u‘) {
            if (lo.empty() && up.empty()) {
                puts("Invalid");
                continue;
            }
            if (cmd[1] == ‘o‘) {
                int x = sta.top();
                sta.pop();
                cout << x << endl;
                pop(x);
            } else if (cmd[1] == ‘e‘) {
                cout << peek() << endl;
            }
        } else {
            cin >> x;
            sta.push(x);
            push(x);
        }
    }
    return 0;
}

此题正解树状数组

#include<stdio.h>  
#include<cstring>  
#include<iostream>  
#include<string>  
using namespace std;  
  
const int N=100005;  
int c[N];  
  
int lowbit(int i){  
    return i&(-i);  
}  
  
void add(int pos,int value){  
    while(pos<N){  
        c[pos]+=value;  
        pos+=lowbit(pos);  
    }  
}  
  
int sum(int pos){  
    int res=0;  
    while(pos>0){  
        res+=c[pos];  
        pos-=lowbit(pos);  
    }  
    return res;  
}  
  
int find(int value){  
    int l=0,r=N-1,median,res;  
    while(l<r-1){  
        if((l+r)%2==0)  
            median=(l+r)/2;  
        else  
            median=(l+r-1)/2;  
        res=sum(median);  
        if(res<value)  
            l=median;  
        else   
            r=median;  
          
    }  
    return l+1;  
}  
  
int main(){  
    //freopen("D://test.txt","r",stdin);  
    char ss[20];  
    int stack[N],top=0,n,pos;  
    memset(c,0,sizeof(c));  
    scanf("%d",&n);  
    while(n--){  
        scanf("%s",ss);  
        if(ss[1]==u){  
            scanf("%d",&pos);  
            stack[++top]=pos;  
            add(pos,1);  
        }else if(ss[1]==o){  
            if(top==0){  
                printf("Invalid\n");  
                continue;  
            }  
            int out=stack[top];  
            add(out,-1);  
            printf("%d\n",stack[top--]);  
        }else if(ss[1]==e){  
            if(top==0){  
                printf("Invalid\n");  
                continue;  
            }  
            int res;  
            if(top%2==0)  
                res=find(top/2);  
            else  
                res=find((top+1)/2);  
            printf("%d\n",res);  
              
        }else{  
            printf("Invalid\n");  
        }  
    }  
  
    return 0;  
}  



类似题目:zoj3612

以上是关于pat1057 stack的主要内容,如果未能解决你的问题,请参考以下文章

PAT (Advanced Level) Practice 1057 Stack

PAT (Advanced Level) 1057 Stack

PAT.1057 Stack(线段树)

PAT.1057 Stack(线段树)

pat1057 stack

PAT 1057 Stack