Missile: Dual-state DP, missile state dp

Missile: Dual-state DP, missile state dp
Question

Description

Long, long ago, country A specified Ted a missile system to destroy the missiles from their enemy. that system can launch only one missile to destroy multiple missiles if the heights of all the missiles form a non-decrease sequence.

But recently, the scientists found that the system is not strong enough. so they invent another missile system. the new system can launch one single missile to destroy more enemy missiles. basically, the system can destroy the missile from near to far. when the system is begun, it chooses one enemy missile to destroy, and then destroys a missile whose height is lower and farther than t He first missile. The third missile to destroy is higher and farther than the second missile... The odd missile to destroy is higher and farther than the previous one, and the even missile to destroy is lower and farther than the previous one.

Now, given you a list of the height of missiles from near to far, please find the most missiles that can be destroyed by one missile launched by the new system.

Input

The input contains multiple test cases.

In each test case, first line is an integer n (0 <n <= 1000), which is the number of missiles to destroy. then follows one line which contains n integers (<= 109), the height of the missiles followed by distance.

The input is terminated by n = 0.

Output

For each case, print the most missiles that can be destroyed in one line.

Sample Input

4
5 3 2 4
3
1 1 1
0

Sample output

3
1

A missile can destroy enemy missiles. However, there is a strange setting:
When an enemy missile comes from far and near, it can choose one of them to start to be destroyed, and then destroy a missile farther and at a lower height than the previous one, then they will destroy missiles that are farther but higher in height.
Calculate the maximum number of missiles.
The first thought of this question is the monotonically increasing sub-sequence deformation. That is, the longest concave-convex subsequence ..
It is obviously a dp problem.
The key to solving the dp problem is to sort out the state transition equation. For monotonically increasing subsequences, the state transition equation is: dp [I] = max (dp [k] + 1) (k <I & v [k] <v [I]);
That is, the longest monotonic subsequence ending with the current vertex is prior to this vertex; the value is smaller than this vertex; and the local optimum + 1 among them.
Because the problem is not monotonic increasing, we need to deform the dp equation.
It is assumed that the rule of missiles is high/low switching. Therefore, each missile may be shot down as a lower or as a high missile. Being shot down as a different identity may have different optimal solutions, and will also affect the subsequent missiles.
Therefore, we need to save two States. The simple convention is: a [I]: Missile I is the optimal solution when higher is shot down. B [I]: the missile is the optimal solution when lower is shot down.
Relatively, the new dp equation is: a [I] = max {B [k] + 1} (k <I & v [k] <v [I]) B [I] = max {a [k] + 1} (k <I & v [k]> v [I])
Initialize a [0] = 1 // The first hit I = 0 to get the 1 solution B [0] = 0 // because the first hit in the question is high, therefore, a missile at the position 0 cannot be shot down as a lower. As the lower, the optimal solution is 0.

Complete code: the code written earlier is not standardized. Let's take a look.

#include<iostream>using namespace std;int a[1001],b[1001];int str[1001];int main(){    //freopen("aa.txt","r",stdin);    int n,i,j,max2;    while(scanf("%d",&n)!=EOF)    {        if(!n)            break;        for(i=0;i<n;i++)            scanf("%d",&str[i]);        for(i=0;i<n;i++)        {            a[i]=1;  //as higher            b[i]=0;  //as lower        }        for(i=1;i<n;i++)        {            int max=1,max1=0;            for(j=0;j<i;j++)            {                if(str[i]>str[j])                {                    a[i]=b[j]+1;                    if(a[i]>max)                        max=a[i];                }                if(str[i]<str[j])                {                    b[i]=a[j]+1;                    if(b[i]>max1)                        max1=b[i];                }            }            b[i]=max1;            a[i]=max;        }        max2=a[0];        for(i=0;i<n;i++)        {            if(a[i]>max2)                max2=a[i];            if(b[i]>max2)                max2=b[i];        }        cout<<max2<<endl;    }}

Verify why the program ensures that the optimal solution is "first hit higher? It can be proved that the default B [I] is 0 because of all positions I. If the optimal solution appears at the lower position, that is, B [k] is the maximum value of all a [] B, it indicates that a [j]> B [k] must exist before k. that is to say, there is a higher first hit.
Simple test.





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