Codeforces Round #358 (Div. 2) Alyona and Strings

D. Alyona and Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

题意:

给出两个字符串 s[1]s[2].......s[n], t[1]t[2]........t[n]

找出k个s的不相交子串,他们同时也依次是t的不相交子串 ,求这些字串长度之和最大值.

Output

In the only line print the only non-negative integer — the sum of the lengths of the strings in a desired sequence.

It is guaranteed, that at least one desired sequence exists.

Examples

input

3 2 2abcab

output

2

input

9 12 4bbaaababbabbbabbaaaba

output

7

Note

The following image describes the answer for the second sample case:

分析：f[k][i][j]表示s[1].....s[i] 与t[1].....t[j] 中选出k个子串，并且s[i]与t[j]一定是第k个子串的结尾的长度和最大值。方程：f[i][j]= -inf (s[i]!=t[j])

f[i][j] =max{

        f[k][i][j]= max{f[k-1][p][q]}+1        k-1<=p<=i ,k-1<=q<=j,  s[i]==t[j]   //s[i]单独一段
f[k][i][j]=f[k][i-1][j-1]+1      s[i]==t[j]&&s[i-1]==t[j-1];    //s[i]与前面的一起

}

k一定的情况下，max{f[k-1][p][q]}其实表示的是f数组中一块矩形中的最大值

记best[i][j]=max{f[k-1][p][q]}

那么best[i-1][j-1]=max{    best[i-1][j-2]);
best[i-2][j-1]);
       f[k-1][i-1][j-1]);

}  （二维前缀和思想）

时间复杂度O（n*m*k）

代码如下：

#include<cstdio>  #include<iostream> #include<cstring> using namespace std;  const int maxn=1000+5,inf=1e9;  int n,m,K,f[11][maxn][maxn],best[maxn][maxn],ans=0;  char s[maxn],t[maxn];int main(){//freopen("data.in","r",stdin);//freopen("myans.out","w",stdout);int i,j,k;scanf("%d%d%d",&n,&m,&K);scanf("%s%s",s+1,t+1);for(k=0;k<=K;k++)for(i=0;i<=n;i++)for(j=0;j<=m;j++)f[k][i][j]=-inf;for(k=1;k<=K;k++){memset(best,0,sizeof(best));for(i=k;i<=n;i++){best[k-1][k-1]=max(f[k-1][k-1][k-1],0);for(j=k;j<=m;j++){if(j>1)best[i-1][j-1]=max(best[i-1][j-1],best[i-1][j-2]);if(i>1)best[i-1][j-1]=max(best[i-1][j-1],best[i-2][j-1]);best[i-1][j-1]=max(best[i-1][j-1],f[k-1][i-1][j-1]);if(s[i]==t[j]) f[k][i][j]=max(f[k][i][j],best[i-1][j-1]+1);    if(s[i]==t[j]&&s[i-1]==t[j-1])f[k][i][j]=max(f[k][i][j],1+f[k][i-1][j-1]);}}}for(i=1;i<=n;i++)for(j=1;j<=m;j++)ans=max(ans,f[K][i][j]);printf("%d\n",ans);return 0;}