标签：递推，DP
Ivan hasan array consisting of n differentintegers. He decided to reorder all elements in increasing order. Ivan lovesmerge sort so he decided to represent his array with one or several increasingsequences which he then plans to merge into one sorted array.
Ivanrepresent his array with increasing sequences with help of the followingalgorithm.
Whilethere is at least one unused number in array Ivan repeats the followingprocedure:
· iterate through array from the left to theright;
· Ivan only looks at unused numbers on currentiteration;
· if current number is the first unused numberon this iteration or this number is greater than previous unused number oncurrent iteration, then Ivan marks the number as used and writes it down.
Forexample, if Ivan's array looks like [1, 3,2, 5, 4] then he will perform two iterations. On first iterationIvan will use and write numbers [1, 3, 5], and onsecond one — [2, 4].
Write aprogram which helps Ivan and finds representation of the given array with oneor several increasing sequences in accordance with algorithm described above.
Input
The firstline contains a single integer n (1 ≤ n ≤ 2·105) — the number ofelements in Ivan's array.
The secondline contains a sequence consisting of distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — Ivan'sarray.
Output
Printrepresentation of the given array in the form of one or more increasingsequences in accordance with the algorithm described above. Each sequence mustbe printed on a new line.
Example
Input
5
1 3 2 5 4
Output
1 3 5
2 4
Input
4
4 3 2 1
Output
4
3
2
1
Input
4
10 30 50 101
Output
10 30 50 101
类似于noip1999拦截导弹那题
为了防止TLE，只好先前加入判断了，这样可以避免一部分的双重循环，时间复杂度O(n*cnt)
//cnt代表输出答案的行数
Code
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline int read()
{
intf=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
returnx*f;
}
struct data{int num,x;}a[200005];
int n,cnt=1,Max[200005];
inline int cmp(data a,data b)
{
if(a.num==b.num)return a.x<b.x;
elsereturn a.num<b.num;
}
int main()
{
n=read();
rep(i,1,n)a[i].x=read();
rep(i,1,n){
if(a[i].x<Max[cnt]){Max[++cnt]=a[i].x;a[i].num=cnt;continue;}
rep(j,1,cnt)if(a[i].x>Max[j]){Max[j]=a[i].x;a[i].num=j;break;}
}
sort(a+1,a+1+n,cmp);
printf("%d",a[1].x);
rep(i,2,n){
if(a[i].num!=a[i-1].num)printf("\n");
printf("%d",a[i].x);
}
return0;
}
