Codeforces Round #484 (Div. 2) B. Bus of Characters

B. Bus of Characters

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In the Bus of Characters there are $\mathrm{nn rows\; of\; seat,\; each\; having 22 seats.\; The\; width\; of\; both\; seats\; in\; the ii-th\; row\; is wiwi centimeters.\; All\; integers wiwi are\; distinct.}$

Initially the bus is empty. On each of $\mathrm{2n2n stops\; one\; passenger\; enters\; the\; bus.\; There\; are\; two\; types\; of\; passengers:}$

• an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it;
• an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it.

You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 2000001\le n\le 200000)\; —\; the\; number\; of\; rows\; in\; the\; bus.}$

The second line contains the sequence of integers ${\mathrm{w1,w2,\dots ,wnw1,w2,\dots ,wn (1\le wi\le 1091\le wi\le 109),\; where wiwi is\; the\; width\; of\; each\; of\; the\; seats\; in\; the ii-th\; row.\; It\; is\; guaranteed\; that\; all wiwi are distinct.}}^{}$

The third line contains a string of length $\mathrm{2n2n,\; consisting\; of\; digits\; \text{'}0\text{'}\; and\; \text{'}1\text{'}\; —\; the\; description\; of\; the\; order\; the\; passengers\; enter\; the\; bus.\; If\; the jj-th\; character\; is\; \text{'}0\text{'},\; then\; the\; passenger\; that\; enters\; the\; bus\; on\; the jj-th\; stop\; is\; an\; introvert.\; If\; the jj-th\; character\; is\; \text{'}1\text{'},\; the\; the\; passenger\; that\; enters\; the\; bus\; on\; the jj-th\; stop\; is\; an\; extrovert.\; It\; is\; guaranteed\; that\; the\; number\; of\; extroverts equals the\; number\; of\; introverts\; (i. e.\; both\; numbers\; equal nn),\; and\; for\; each\; extrovert\; there always is\; a\; suitable\; row.}$

Output

Print $\mathrm{2n2n integers\; —\; the\; rows\; the\; passengers\; will\; take.\; The\; order\; of\; passengers\; should\; be\; the\; same\; as\; in\; input.}$

Examples

input

Copy

2
3 1
0011

output

Copy

2 1 1 2 

input

Copy

6
10 8 9 11 13 5
010010011101

output

Copy

6 6 2 3 3 1 4 4 1 2 5 5 

Note

In the first example the first passenger (introvert) chooses the row $\mathrm{22,\; because\; it\; has\; the\; seats\; with\; smallest\; width.\; The\; second\; passenger\; (introvert)\; chooses\; the\; row 11,\; because\; it\; is\; the\; only\; empty\; row\; now.\; The\; third\; passenger\; (extrovert)\; chooses\; the\; row 11,\; because\; it\; has\; exactly\; one\; occupied\; seat\; and\; the\; seat\; width\; is\; the\; largest\; among\; such\; rows.\; The\; fourth\; passenger\; (extrovert)\; chooses\; the\; row 22,\; because\; it\; is\; the\; only\; row\; with\; an\; empty\; place.}$

附上代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 2*1e5+5;
typedef struct NODE
{
    int math;
    int flag;
    friend bool operator < (NODE a, NODE b)
    {
        return a.math > b.math;    //重载小于号使得小的先出队列
    }
} node;
node team[MAXN];
node team1[MAXN];
int bj[MAXN];
int people[MAXN*2];
priority_queue<node>que1;///保存大的值
priority_queue<node>que2;///保存小的值
priority_queue<node>que3;///保存结果
priority_queue<node>que4;///保存另一个结果
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(team,0,sizeof(team));
        memset(bj,0,sizeof(bj));
        memset(people,0,sizeof(people));
        while(!que3.empty())
        {
            que3.pop();
        }
        while(!que1.empty())
        {
            que1.pop();
        }
        while(!que2.empty())
        {
            que2.pop();
        }
        while(!que4.empty())
        {
            que4.pop();
        }
        for(int i=0; i<n; i++)
        {
            scanf("%d",&team[i].math);
            team1[i].math=-team[i].math;
            team1[i].flag=i;
            team[i].flag=i;
            que1.push(team1[i]);
            que2.push(team[i]);
        }
        char str[MAXN*2];
        scanf("%s",str);
        for(int i=0; i<n*2; i++)
        {
            int ans;
            ans=str[i]-'0';
            if(ans==0)
            {//printf("!!!!\n");
                if(!que1.empty())
                {
                    node ans1=que2.top();
                    //printf("%d\n",ans1.math);
                    que2.pop();
                    if(bj[ans1.flag]==0)
                    {
                        que3.push(ans1);
                        bj[ans1.flag]=1;
                        people[i]=ans1.flag+1;
                        ans1.math=-ans1.math;
                        que4.push(ans1);
                    }
                }
            }
            if(ans==1)
            {//printf("..?.\n");
                if(!que4.empty())
                {
                    node ans2;
                    ans2=que4.top();
                    que4.pop();
                    people[i]=ans2.flag+1;
                }
                else
                    if(que4.empty())
                {
                    node ans3;
                    ans3=que1.top();
                    que1.pop();
                    bj[ans3.flag]=1;
                    people[i]=ans3.flag+1;
                }
            }
        }
        for(int i=0;i<2*n;i++)
            printf("%d ",people[i]);
        puts("");
    }
    return 0;
}