http://acm.hdu.edu.cn/showproblem.php?pid=5493
QueueTime 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. Input The first line of input contains a number T indicating the number of test cases (T≤1000 ). 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;
}