POJ 1423 Greatest Common Increasing Subsequence【裸LCIS】

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1423

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28195#problem/F

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2757    Accepted Submission(s): 855

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

151 4 2 5 -124-12 1 2 4

 

Sample Output

2

 

Source

ACM暑期集训队练习赛（二）

 

Recommend

lcy

算法：

 

LCIS 【最长公共上升子序列分析】

code：

注意格式 问题：

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int maxn = 500+50;int dp[maxn][maxn];int a[maxn],b[maxn];int m,n;/****求序列 A 长度为 N 和序列 B 长度为 M 的 LCS序列下标从 1 开始*/int LCS(){    for(int i = 1; i <= n; i++)    {        int tmp = 0; //记录在i确定，且a[i]>b[j]的时候dp[i,j]的最大值        for(int j = 1; j <= m; j++)        {            dp[i][j] = dp[i-1][j];            if(a[i] > b[j])            {                tmp = dp[i-1][j];            }            else if(a[i] == b[j])                dp[i][j] = tmp+1;        }    }//for(int i = 1; i <= m; i++) printf("%d ", dp[n][i]); printf("\n");    int ans = 0;    for(int i = 1; i <= m; i++)        ans = max(ans, dp[n][i]);    return ans;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        memset(dp,0,sizeof(dp));        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]);        scanf("%d", &m);        for(int j = 1; j <= m; j++)            scanf("%d", &b[j]);        printf("%d\n",LCS());        if(T != 0) printf("\n");    }}

内存优化：

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int maxn = 500+50;int dp[maxn];int a[maxn],b[maxn];int m,n;/****求序列 A 长度为 N 和序列 B 长度为 M 的 LCS序列下标从 1 开始*/int LCS(){    for(int i = 1; i <= n; i++)    {        int tmp = 0;        for(int j = 1; j <= m; j++)        {            if(a[i] > b[j] && dp[j] > tmp)            {                tmp = dp[j];            }            else if(a[i] == b[j])                dp[j] = tmp+1;        }    }    int ans = 0;    for(int i = 1; i <= m; i++)        ans = max(ans, dp[i]);    return ans;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        memset(dp,0,sizeof(dp));        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]);        scanf("%d", &m);        for(int j = 1; j <= m; j++)            scanf("%d", &b[j]);        printf("%d\n",LCS());        if(T != 0) printf("\n");    }}