HDU 5493 Queue（树状数组+二分）

灰太狼 2022-06-08 02:59 174阅读 0赞

N N people numbered from 1 to N N are waiting in a bank for service. They all stand in a queue, but the queue never moves. It is lunch time now, so they decide to go out and have lunch first. When they get back, they don’t remember the exact order of the queue. Fortunately, there are some clues that may help.  
Every person has a unique height, and we denote the height of the i i\-th person as hi hi. The i i\-th person remembers that there were ki ki people who stand before him and are taller than him. Ideally, this is enough to determine the original order of the queue uniquely. However, as they were waiting for too long, some of them get dizzy and counted ki ki in a wrong direction. ki ki could be either the number of taller people before or after the i i\-th person.  
Can you help them to determine the original order of the queue?

Input

The first line of input contains a number T T indicating the number of test cases ( T≤1000 T≤1000).  
Each test case starts with a line containing an integer N N indicating the number of people in the queue ( 1≤N≤100000 1≤N≤100000). Each of the next N N lines consists of two integers hi hi and ki ki as described above ( 1≤hi≤109,0≤ki≤N−1 1≤hi≤109,0≤ki≤N−1). Note that the order of the given hi hi and ki ki is randomly shuffled.  
The sum of N N over all test cases will not exceed 106 106

Output

For each test case, output a single line consisting of “Case \#X: S”. X X is the test case number starting from 1. S S is people’s heights in the restored queue, separated by spaces. The solution may not be unique, so you only need to output the smallest one in lexicographical order. If it is impossible to restore the queue, you should output “impossible” instead.

Sample Input

3
    3
    10 1
    20 1
    30 0
    3
    10 0
    20 1
    30 0
    3
    10 0
    20 0
    30 1

Sample Output

Case #1: 20 10 30
    Case #2: 10 20 30
    Case #3: impossible

题解：

题意：

给你人的身高，和比他高的人数，输出字典序最小的可能排列，否则输出impossible

思路：

先将人身高从小到大排序，他在队列中的位置可以是pos或者n-pos，然后二分寻找符合条件的位置更新就行了

代码：

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    #include<algorithm>
    #define ll long long
    using namespace std;
    const int maxn=1e5+5;
    struct node
    {
        int h,pos;
    }a[maxn];
    int b[maxn];
    int p[maxn];
    int cmp(node x,node y)
    {
        return x.h<y.h;
    }
    int lowbit(int x)
    {
        return x&(-x);
    }
    void update(int x,int v)
    {
        while(x<maxn)
        {
            b[x]+=v;
            x+=lowbit(x);
        }
    }
    int query(int x)
    {
        int s=0;
        while(x>0)
        {
            s+=b[x];
            x-=lowbit(x);
        }
        return s;
    }
    int main()
    {
        int i,j,k,n,m,q,test,tag,l,r,mid;
        scanf("%d",&test);
        for(q=1;q<=test;q++)
        {
            scanf("%d",&n);
            memset(b,0,sizeof(b));
            for(i=1;i<=n;i++)
            {
                scanf("%d%d",&a[i].h,&a[i].pos);
            }
            sort(a+1,a+n+1,cmp);
            tag=0;
            for(i=1;i<=n;i++)
            {
                k=min(a[i].pos,n-i-a[i].pos);
                if(k<0)
                {
                    tag=1;
                    break;
                }
                k++;
                l=1,r=n;
                while(l<r)
                {
                    mid=(l+r)/2;
                    if(mid-query(mid)>=k)
                        r=mid;
                    else
                        l=mid+1;
                }
                update(l,1);
                p[l]=a[i].h;
            }
            printf("Case #%d: ",q);
            if(tag)
                printf("impossible");
            else
            {
                printf("%d",p[1]);
                for(i=2;i<=n;i++)
                {
                    printf(" %d",p[i]);
                }
            }
            printf("\n");
        }
        return 0;
    }