HDU5493-Queue

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Queue

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1177 Accepted Submission(s): 608

Problem Description

N people numbered from 1 to

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-th person as

hi. The

i-th person remembers that there were

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 in a wrong direction.

ki could be either the number of taller people before or after the

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 indicating the number of test cases (

T≤1000).
Each test case starts with a line containing an integer

N indicating the number of people in the queue (

1≤N≤100000). Each of the next

N lines consists of two integers

hi and

ki as described above (

1≤hi≤109,0≤ki≤N−1). Note that the order of the given

hi and

ki is randomly shuffled.
The sum of

N over all test cases will not exceed

106

Output

For each test case, output a single line consisting of “Case #X: S”.

X is the test case number starting from 1.

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

3310 120 130 0310 020 130 0310 020 030 1

Sample Output

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

Source

2015 ACM/ICPC Asia Regional Hefei Online

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题意：给定人的身高，和排在他前面或后面比他高的人的个数。构造一个序列使得满足，并且字典序最小。

解题思路：二分+树状数组，可以先按照身高从小到大排序，假设当前到了第i高的人，他前面或者后面有k个人，那么他前面的所有人都比他矮，比他高的还有n-i个人，因为要求字典序最小，所以他在未排的人中的位置 p = min(k, n - i - k)。每个人有两个可能位置，因为他之前的都比他矮，那么为了让字典序最小，就选择一个较小的位置。当n - i - k < 0 时，说明没有多余空格，那么无解。二分第i个人的位置，用树状数组判断他前面有多少人以此确定空余的位置数。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;struct node{int h, k;}a[100009];int c[100009];int ans[100009];int n;int lowbit(int x){return x&-x;}void add(int x,int val){while (x <= n){c[x]+=val;x += lowbit(x);}}int getsum(int x){int sum = 0;while (x){sum += c[x];x -= lowbit(x);}return sum;}bool cmp(node a, node b){return a.h < b.h;}int main(){int t, cas = 0;scanf("%d", &t);while (t--){scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].h, &a[i].k);memset(c, 0, sizeof c);sort(a + 1, a + 1 + n, cmp);for (int i = 1; i <= n; i++) add(i,1);int flag = 1;for (int i = 1; i <= n; i++){int k = n - i,kk=a[i].k;if (kk > k) { flag = 0; break; }int temp = min(kk+1,k-kk+1);int l = 1, r = n,ans1;while (l <= r){int mid = (l + r) >> 1;if (getsum(mid) >= temp) { ans1 = mid, r = mid - 1; }else l = mid + 1;}ans[ans1] = a[i].h;add(ans1, -1);}if (!flag) printf("Case #%d: impossible\n",++cas);else{printf("Case #%d:", ++cas);for (int i = 1; i <= n; i++) printf(" %d", ans[i]);printf("\n");}}return 0;}

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