2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest, qualification stage (Online Mirror, ACM-ICP

补B题:
链接:http://codeforces.com/contest/847/problem/B

Ivan has an array consisting of n different integers. He decided to reorder all elements in increasing order. Ivan loves merge sort so he decided to represent his array with one or several increasing sequences which he then plans to merge into one sorted array.

Ivan represent his array with increasing sequences with help of the following algorithm.

While there is at least one unused number in array Ivan repeats the following procedure:

iterate through array from the left to the right;
Ivan only looks at unused numbers on current iteration;
if current number is the first unused number on this iteration or this number is greater than previous unused number on current iteration, then Ivan marks the number as used and writes it down.
For example, if Ivan’s array looks like [1, 3, 2, 5, 4] then he will perform two iterations. On first iteration Ivan will use and write numbers [1, 3, 5], and on second one — [2, 4].

Write a program which helps Ivan and finds representation of the given array with one or several increasing sequences in accordance with algorithm described above.

Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of elements in Ivan’s array.

The second line contains a sequence consisting of distinct integers a1, a2, …, an (1 ≤ ai ≤ 109) — Ivan’s array.

Output
Print representation of the given array in the form of one or more increasing sequences in accordance with the algorithm described above. Each sequence must be printed on a new line.

Examples

inputCopy
5
1 3 2 5 4
outputCopy
1 3 5
2 4

inputCopy
4
4 3 2 1
outputCopy
4
3
2
1

inputCopy
4
10 30 50 101
outputCopy
10 30 50 101

题意:给你n个数字,然后你输出k个严格上升的序列;例如n = 5,分别是5 7 5 2 6
那么 第一个序列首先放进5,k1[5]
第二个数字7比第一个序列的尾部大,所以放进第一个序列的尾部,[5,7]
第三个数字5比第一个序列的尾部小,所以新开一个序列k2并放到k2,k2[5]
第4个数字是2,比k1尾部大,再来也比k2尾部大,所以新开k3并放到k3,k3[2]
第5个数字是6,比k1尾部小,再来比k2尾部大,所以放到k2尾部;k2[5,6]
输出k个序列;[5,7]
[5,6]
[2]

做法:我们不难发现,k个序列的长度是有规律的,Numbe数目 k >= Number k+i;而且
只看k序列的最后一个,那么,它也是非严格递减的;
所以,先判断 r(最后一个序列的尾部),如果a[i] < r,直接插入 k+1 的那个序列;
a[i] >= r, 二分查找 要插入的位置;

AC_Code:

#include <bits/stdc++.h>      //https://www.runoob.com/w3cnote/cpp-vector-container-analysis.html
using namespace std;

const int maxn = 21e5+10; 	
int a[maxn];
int qlen = 0;
struct node{
	vector<int>b;
}q[maxn];

int main(){
	ios::sync_with_stdio(false);
	int n;
	while(cin >> n){
		qlen = 0;
		for(int i = 0;i < n;i++){
			cin >> a[i];
		}
		q[0].b.push_back(a[0]);qlen = 1;
		for(int i = 1;i < n;i++){
			if(q[qlen-1].b.back()> a[i]){        //can new 
				q[qlen++].b.push_back(a[i]);
				continue;
			}            
			int low = 0,high = qlen;         //can not new er fen
			while(low < high){
				int mid = (low + high)/2;
				if(q[mid].b.back() < a[i]){
					high = mid;
				}else{
					low = mid+1;
				}
			}
			q[low].b.push_back(a[i]);
		}
		//print()	
		for(int i = 0;i < qlen;i++){
			vector<int>::iterator it;    //迭代器,相当于指针	
			for(it = q[i].b.begin();it!=q[i].b.end();it++){
				it == q[i].b.begin()?cout << *it : cout << " " << *it;
			}
			cout << endl;
			q[i].b.clear();
		}
	}
}