HDU 1423 Greatest Common Increasing Subsequence(LICS入门，只要求出最长数)

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5444    Accepted Submission(s): 1755

Problem Description

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

 

Input

Each
sequence is described with M - its length (1 <= M <= 500) and M
integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

 

Output

output print L - the length of the greatest common increasing subsequence of both sequences.

 

Sample Input

1

5
1 4 2 5 -12
4
-12 1 2 4

 

Sample Output

2

 

Source

ACM暑期集训队练习赛（二）

 

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四种方法代码：

每一种都按照我觉得最容易理解的思路和最精简的写法写了，有些地方i 或者 i-1都可以，不必纠结。

#include<algorithm>
#include<string>
#include<string.h>
#include<stdio.h>
#include<iostream>
using namespace std;
#define N 505

int a[N],l1,l2,b[N],ma,f[N][N],d[N];
void solve1()//O(n^4)
{
    ma = ;
    memset(f, , sizeof(f) );
    for(int i = ; i <= l1; i++)
        for(int j = ; j <= l2; j++)
        {
            if(a[i-] == b[j-])
            {
                int ma1 = ;
                for(int i1 = ; i1 < i; i1++)
                    for(int j1 = ; j1 < j; j1++)
                        if(a[i1-] == b[j1-]  &&  a[i1-] < a[i-]  &&  f[i1][j1] > ma1)//实际上a[i-1]==b[j1-1]可以省，因为既然f[i1][j1]+1>f[i][j]了，
                            ma1 = f[i1][j1];                                             /*说明肯定a[i-1]==b[j1-1]了，因为这种解法只有当a[i-1]==b[j-1]时，f[i][j]才不等于0*/
                f[i][j] = ma1 + ;
            }
            ma = max(ma, f[i][j]);
        }
    cout<<ma<<endl;
}
void solve2()//O(n^3)
{
    ma = ;
    memset(f, , sizeof(f) );
    for(int i = ; i <= l1; i++)
        for(int j = ; j <= l2; j++)
        {
            f[i][j] = f[i-][j];
            if(a[i-] == b[j-])
            {
                int ma1 = ;
                for(int j1 = ; j1 < j; j1++)
                {
                    if(b[j1-] < b[j-]  &&  f[i-][j1] > ma1)
                        ma1 = f[i-][j1];
                }
                f[i][j] = ma1 + ;
        }
        ma = max(ma, f[i][j]);
    }
    cout<<ma<<endl;
}
void solve3()//O(n^2)
{
    memset(f, , sizeof(f) );
    for(int i = ; i <= l1; i++)
    {
        int  ma1 = ;
        for(int j = ; j <= l2; j++)
        {
            f[i][j] = f[i-][j];//带这种的一般都能压缩一维空间，也就是简化空间复杂度
            if(a[i-] > b[j-]  &&  f[i-][j] > ma1)
                ma1 = f[i-][j];
            if(a[i-] == b[j-])
                f[i][j] = ma1 + ;
        }
    }
    ma = -;
    for(int j = ;j <= l2; j++)
        ma=max(ma,f[l1][j]);

    cout<<ma<<endl;
}
void solve4()//O(n^2)//优化空间复杂度
{
    ma = ;
    memset(d, , sizeof(d) );
    for(int i = ; i <= l1; i++)
    {
        int ma1 = ;
        for(int j = ; j <= l2; j++)
        {
            if(a[i-] > b[j-]  &&  d[j] > ma1)
                ma1 = d[j];
            if(a[i-] == b[j-])
                d[j] = ma1 + ;
        }
    }
    ma = -;
    for(int j = ;j <= l2; j++)
        ma = max(ma, d[j]);

    cout<<ma<<endl;
}

int main()
{
    int T;cin>>T;
    while(T--)
    {
        scanf("%d", &l1);
        for(int i = ; i < l1; i++)
            scanf("%d", &a[i]);
        scanf("%d", &l2);
        for(int i = ; i < l2; i++)
            scanf("%d", &b[i]);
//        solve1();
//        solve2();
//        solve3();
        solve4();

        if(T)
            printf("\n");
    }

    return ;
}
/*
99

5
1 4 2 5 -12
4
-12 1 2 4

9
3 5 1 6 7 9 1 5 13
6
4 6 13 9 13 5
*/