HZAU 17:LCS

17: LCS

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 184  Solved: 43
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Description

Giving two strings consists of only lowercase letters, find the LCS(Longest Common Subsequence) whose all partition are not less than k in length.

Input

There are multiple test cases. In each test case, each of the first two lines is a string(length is less than 2100). The third line is a positive integer k. The input will end by EOF.

Output

For each test case, output the length of the two strings’ LCS.

Sample Input

abxccdef
abcxcdef
3
abccdef
abcdef
3

Sample Output

4
6

题目分析：设两个字符串分别为p和q，定义状态dp(i,j)表示p的前缀p(1~i)和q的前缀q(1~j)的满足题目要求的LCS的长度，定义g(i,j)表示满足题目要求的并且结尾字符为p(i)或q(j)（显然，这要求p(i)=q(j)）的LCS的长度。
的长度。则状态转移方程为：dp(i,j)=max(dp(i-1,j),dp(i,j-1),g(i,j))。

代码如下：

# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
using namespace std;

int k;
char p[2105];
char q[2105];
int f[2105][2105];
int g[2105][2105];
int dp[2105][2105];

int main()
{
    //freopen("F.in","r",stdin);
    while(~scanf("%s%s%d",p+1,q+1,&k))
    {
        memset(f,0,sizeof(f));
        memset(g,0,sizeof(g));
        memset(dp,0,sizeof(dp));
        int n=strlen(p+1);
        int m=strlen(q+1);
        for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
                if(p[i]==q[j]) f[i][j]=f[i-1][j-1]+1;
        for(int i=1;i<=n;++i){
            for(int j=1;j<=m;++j){
                if(f[i][j]>=k){
                    g[i][j]=max(g[i][j],dp[i-k][j-k]+k);
                    if(f[i][j]>k)
                        g[i][j]=max(g[i][j],g[i-1][j-1]+1);
                }
                dp[i][j]=max(max(dp[i-1][j],dp[i][j-1]),g[i][j]);
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}