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): 3079 Accepted Submission(s): 969


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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//0MS    1248K    1342 B    C++
//最长公共递增子序列
//先求出公共子序列再从公共子序列中求递增子序列 
#include<stdio.h>
#include<string.h>
int a[505],b[505];
int dp[505][505];
int Max(int a,int b)
{
    return a>b?a:b;
}
int main(void)
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        int ans[505];//存放公共子序列 
        memset(ans,-1,sizeof(ans));
        for(int i=1;i<=n;i++){
            int temp=0;
            for(int j=1;j<=m;j++){
                if(a[i-1]==b[j-1]){
                    dp[i][j]=Max(dp[i][j],dp[i-1][j-1]+1);
                    if(ans[dp[i][j]]==-1)
                        ans[dp[i][j]]=a[i-1];
                }
                else {
                    dp[i][j]=Max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        int maxn=0;
        int dp0[505]={0};
        for(int i=1;i<=dp[n][m];i++){
            //printf("%d\n",ans[i]);
            int temp=0;
            for(int j=1;j<i;j++)
                if(ans[i]>ans[j] && temp<dp0[j])
                    temp=dp0[j];
            dp0[i]=temp+1;
            if(dp0[i]>maxn) maxn=dp0[i];;
        }
        printf("%d\n",maxn);
        if(t) printf("\n"); //题目没说明错了两次 
    }
    return 0;
}        

 

//0MS    256K    926 B    C++
//借鉴别人的直接解法(模板) 
//直接求最长公共递增子序列数 
#include<stdio.h>
#include<string.h>
int a[505],b[505];
int dp[505]; //记录最长公共递增子序列数 
int main(void)
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        int maxn=0;
        for(int i=0;i<n;i++){
            int temp=0;
            for(int j=0;j<m;j++){
                if(b[j]<a[i] && dp[j]>dp[temp]) temp=j;
                if(a[i]==b[j]) dp[j]=dp[temp]+1;
                //printf("%d %d %d\n",i,j,dp[j]);
            }   
        }
        for(int i=0;i<m;i++)
            if(dp[i]>maxn) maxn=dp[i];
        printf("%d\n",maxn);
        if(t) printf("\n");
    }
    return 0;
}