| Time Limit: 4000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
Description
Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…
The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.
It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!
People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.
Input
There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Valiare as follows:
There no blank lines between test cases. Proceed to the end of input.
Output
For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.
Sample Input
4 0 77 1 51 1 33 2 69 4 0 20523 1 19243 1 3890 0 31492
Sample Output
77 33 69 51 31492 20523 3890 19243
Hint
The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.
题目大意：
n个人站队（可以插队），Pi代表i插队时他前面的人数，Vi代表i的标记，最终输出的是最终队伍的序列。参考代码：
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define lson l,mid,cur<<1
#define rson mid+1,r,cur<<1|1
#define root 1,n,1
const double eps=1e-10;
const int INF=0x3f3f3f3f;
const int MAXN=200000+50;
typedef long long LL;
int n,sum[MAXN<<2],loc[MAXN],num[MAXN],people[MAXN];
void pushup(int cur)
{
sum[cur]=sum[cur<<1]+sum[cur<<1|1];
}
void build(int l,int r,int cur)
{
if(l==r)
{
sum[cur]=1;
return ;
}
int mid=(l+r)/2;
build(lson);
build(rson);
pushup(cur);
}
void update(int a,int b,int l,int r,int cur)
{
if(l==r)
{
people[l]=b;
sum[cur]=0;
return ;
}
int mid=(l+r)/2;
if(sum[cur<<1]>=a)
update(a,b,lson);
else
update(a-sum[cur<<1],b,rson);
pushup(cur);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
while(scanf("%d",&n)!=EOF)
{
build(root);
for(int i=1;i<=n;i++)
scanf("%d%d",&loc[i],&num[i]);
for(int i=n;i>=1;i--)//从后往前依次插队
update(loc[i]+1,num[i],root);//注意loc[i]要加1
for(int i=1;i<=n;i++)
{
if(i!=n)
printf("%d ",people[i]);
else
printf("%d\n",people[i]);
}
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
原文地址：http://blog.csdn.net/noooooorth/article/details/47735229
