hdu 1423 Greatest Common Increasing Subsequence

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2990    Accepted Submission(s): 935

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

最长公共上升子序列（LCIS）

二维做法

 

#include <iostream>
#include <string>
#define maxn 1010
#define max(a, b) a > b ? a : b
using namespace std;
 
int a[maxn], b[maxn];
int dp[2][maxn], ans;
 
int main(){
    int T, n1, n2;
    scanf("%d", &T);
    while ( T-- ){
        memset( dp, 0, sizeof(dp) );
        scanf("%d", &n1);
        for ( int i = 1; i <= n1; i++ ){
            scanf("%d", &a[i]);
        }
        scanf("%d", &n2);
        for ( i = 1; i <= n2; i++ ){
            scanf("%d", &b[i]);
        }
        ans = 0;
        for ( i = 1; i <= n1; i++ ){
            for ( int j = 1; j <= n2; j++ ){
                if ( a[i] == b[j] ){
                    dp[i&1][j] = dp[(i-1)&1][j-1] + 1;
                }
                else if( a[i] > b[j] ){        //只从比自己小的数那里继承最优状态
                    dp[i&1][j] = max( dp[i&1][j-1], dp[(i-1)&1][j] );
                }
                if ( dp[i&1][j] > ans ){
                    ans = dp[i&1][j];
                }
            }
        }
        printf("%d\n", ans);
        if ( T ){
            printf("\n");
        }
    }
}

 

一维的做法

#include <iostream>
#include <string>
#define maxn 1010
//#define max(a, b) a > b ? a : b
using namespace std;
 
int a[maxn], b[maxn];
int dp[maxn], ans;
 
int main(){
    int T, n1, n2;
    scanf("%d", &T);
    while ( T-- ){
        memset( dp, 0, sizeof(dp) );
        scanf("%d", &n1);
        for ( int i = 1; i <= n1; i++ ){
            scanf("%d", &a[i]);
        }
        scanf("%d", &n2);
        for ( i = 1; i <= n2; i++ ){
            scanf("%d", &b[i]);
        }
        ans = 0;
        for ( i = 1; i <= n1; i++ ){
            int max = 0;                                    //保存比a[i]小的当最优状态
            for ( int j = 1; j <= n2; j++ ){
                if ( a[i] == b[j] ){
                    dp[j] = max + 1;
                }
                else if( a[i] > b[j] && dp[j] > max ){        //只从比自己小的数那里继承最优状态
                    max = dp[j];
                }
                if ( dp[j] > ans ){
                    ans = dp[j];
                }
            }
        }
        printf("%d\n", ans);
        if ( T ){
            printf("\n");
        }
    }
    return 0;
}

 

来自为知笔记(Wiz)