hiho一下 第六十周 题目1 : String Matching Content Length dp 最长公共子序列

题目1 : String Matching Content Length

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

We define the matching contents in the strings of strA and strB as common substrings of the two strings. There are two additional restrictions on the common substrings.

The first restriction here is that every common substring's length should not be less than 3.  For example:

strA: abcdefghijklmn
strB: ababceghjklmn

The matching contents in strA and strB are substrings ("abc", "jklmn"). Note that though "e" and "gh" are common substrings of strA and strB, they are not matching content because their lengths are less than 3.

The second restriction is that the start indexes of all common substrings should be monotone increasing. For example:

strA: aaabbbbccc
strB: aaacccbbbb

The matching contents in strA and strB are substrings ("aaa", "bbbb"). Note that though "ccc" is common substring of strA and strB and has length not less than 3, the start indexes of ("aaa", "bbbb", "ccc") in strB are (0, 6, 3), which is not monotone increasing.

输入

Two lines. The first line is strA and the second line is strB. Both strA and strB are of length less than 2100.

输出

The maximum length of matching contents (the sum of the lengths of the common substrings).

样例输入

abcdefghijklmnababceghjklmn

样例输出

8

题意，要求最大的公共子序列，要求共公子序列的每一个子串都要是大于3的

类似于经典的最长公共子序列的做法。

dp[i][j][3]第一个串s1前i位  第二个串s2前j位，最后一个子串长度为0 1 2 3(>=3)的最长公共子序列。

状态转移

1,dp[i][j][0]转移到dp[i-1][j][0],dp[i-1][j][3] dp[i][j-1][0] dp[i][j-1][3];也就是第s1[i] 与s2[j]不匹配，直接转移到前一位就可以了。

2.s1[i] == s2[j]则，dp[i][j][k]可以转移到dp[i][j][k-1];也就是同是加一位。

初始化的时候，要注意，dp[0][i],dp[i][0]都为0;

由于有n * n个状态，状态转移为o(1)所有总的复杂度为O(n *n);

其次，给些测试数据

abcdefghijklmn
ababceghjklmn


aaabbbbccc
aaacccbbbb


aaa
aaa
ab
ab
aaaab
baaaa


abcba
bcbaa


babad
babacabad


abcdef
abcxcdef

#define N 2205#define M 100005#define maxn 2000005#define MOD 1000000000000000007int n,m,dp[N][N][4];char s1[N],s2[N];int main(){    //freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);     while(SS(s1)!=EOF)    {        SS(s2);        n = strlen(s1);m = strlen(s2);        FI(n) FJ(m) For(k,0,4) dp[i][j][k] = -1;        FI(n+1)            dp[0][i][0] = dp[i][0][0] =0;        For(i,1,n+1){            For(j,1,m+1){                dp[i][j][0] = max(dp[i][j][0],dp[i][j-1][0]);                dp[i][j][0] = max(dp[i][j][0],dp[i-1][j][0]);                dp[i][j][0] = max(dp[i][j][0],dp[i][j-1][3]);                dp[i][j][0] = max(dp[i][j][0],dp[i-1][j][3]);                if(s1[i-1] == s2[j-1]){                    For(k,1,4)                        if(dp[i-1][j-1][k-1] >= 0)                            dp[i][j][k] = max(dp[i][j][k],dp[i-1][j-1][k-1] + 1);                    if(dp[i-1][j-1][3] >= 0)                        dp[i][j][3] = max(dp[i][j][3],dp[i-1][j-1][3] + 1);                }            }        }        printf("%d\n",max(0,max(dp[n][m][0],dp[n][m][3])));    }    //fclose(stdin);    //fclose(stdout);    return 0;}