HDU5493 Queue二分+树状数组

http://acm.hdu.edu.cn/showproblem.php?pid=5493

Queue Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1570    Accepted Submission(s): 808   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 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

#include<stdio.h>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int MAXN=200000+5;//最大元素个数
int n;//元素个数
int c[MAXN],ans[MAXN];//c[i]==A[i]+A[i-1]+...+A[i-lowbit(i)+1]
//返回i的二进制最右边1的值
int lowbit(int i)
{
    return i&(-i);
}
//返回A[1]+...A[i]的和
int sum(int x){
    int sum = 0;
    while(x){
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}
//令A[i] += val
void add(int x, int val){
	
    while(x <= n){
        c[x] += val;
        x += lowbit(x);
    }
}
int find_(int x)
{
	int l=1,r=n,mid;
	while(l<r)
	{
		 mid=(l+r)>>1;
		int num=sum(mid);
		if(num<x)
			l=mid+1;
		else
			r=mid;
	}
	return l;
}
struct node1
{
	ll h,v;
}a[MAXN];
ll b[MAXN];
bool cmp(node1 x,node1 y)
{
	 return x.h<y.h;
}
int main()
{
   
   int t;
    t--;
    cin>>t;
    int kase=0;
    while(t--)
	{
		kase++;
	  memset(c,0,sizeof(c));
	  memset(ans,0,sizeof(ans));
	  scanf("%lld",&n);
	  for(int i=1;i<=n;i++)
	  {
	  	add(i,1);
	  	scanf("%lld%lld",&a[i].h,&a[i].v);
	  } 
	  sort(a+1,a+n+1,cmp);
	  int flag=1;
	 for(int i=1;i<=n;i++)
	 {
	 	if(n-i-a[i].v<0) {flag=0;break;}
        int p=min(a[i].v,n-i-a[i].v)+1; ///前面有多少个空位，包括本身，所以+1
		int pos=find_(p);
		
	  	add(pos,-1);
	  	ans[pos]=a[i].h;
	  }
	  printf("Case #%d:", kase);
		if (flag) 
		{
			for (int i = 1; i <= n; i++) {
				printf(" %d", ans[i]);
			}
			printf("\n");
		}
		else 
			printf(" impossible\n");
	
	}
    return 0;
}