CF 346B. Lucky Common Subsequence(DP+KMP)

这题确实很棒。。又是无想法。。其实是AC自动机+DP的感觉，但是只有一个串，用kmp就行了。

dp[i][j][k]，k代表前缀为virus[k]的状态，len表示其他所有状态串，处理出Ac[len][26]数组来，DP就可以了。状态转移那里一直没想清楚，wa了很多次，记录路径倒是不复杂，瞎搞搞就行。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
char s1[103],s2[103],virus[103];
int dp[103][103][103];
int pre[103][103][103];
int pre1[103][103][103];
int Ac[111][31];
int next[103];
char ans[111];
void kmp()
{
    int i,j,len,temp;
    len = strlen(virus);
    next[0] = -1;
    j = -1;
    for(i = 1; i < len; i ++)
    {
        while(j >= 0&&virus[j+1] != virus[i])
            j = next[j];
        if(virus[j+1] == virus[i]) j ++;
        next[i] = j;
    }
    for(i = 0; i < len; i ++)
    {
        for(j = 0; j < 26; j ++)
        {
            temp = i;
            while(temp >= 0&&virus[temp+1] != 'A'+j)
                temp = next[temp];
            if(virus[temp+1] == 'A' + j) temp ++;
            if(temp == -1)
                Ac[i][j] = len;
            else
                Ac[i][j] = temp;
        }
    }
    for(i = 0; i < 26; i ++)
    {
        if(i + 'A' == virus[0])
            Ac[len][i] = 0;
        else
            Ac[len][i] = len;
    }
}
int main()
{
    int i,j,k,len1,len2,len,maxz,a,b,kk;
    scanf("%s%s%s",s1,s2,virus);
    len1 = strlen(s1);
    len2 = strlen(s2);
    len = strlen(virus);
    kmp();
    for(i = 1; i <= len1; i ++)
    {
        for(j = 1; j <= len2; j ++)
        {
            for(k = 0; k <= len; k ++)
            {
                if(k == len-1) continue;
                if(dp[i][j][k] < dp[i-1][j][k])
                {
                    dp[i][j][k] = dp[i-1][j][k];
                    pre[i][j][k] = 2;
                    pre1[i][j][k] = k;
                }
                if(dp[i][j][k] < dp[i][j-1][k])
                {
                    dp[i][j][k] = dp[i][j-1][k];
                    pre[i][j][k] = 3;
                    pre1[i][j][k] = k;
                }
                if(s1[i-1] == s2[j-1])
                {
                    if(Ac[k][s1[i-1]-'A'] == len-1) continue;
                    else if(dp[i][j][Ac[k][s1[i-1]-'A']] < dp[i-1][j-1][k] + 1)
                    {
                        dp[i][j][Ac[k][s1[i-1]-'A']] = dp[i-1][j-1][k] + 1;
                        pre[i][j][Ac[k][s1[i-1]-'A']] = 1;
                        pre1[i][j][Ac[k][s1[i-1]-'A']] = k;
                    }
                }
            }
        }
    }
    maxz = 0;
    for(i = 1; i <= len1; i ++)
    {
        for(j = 1; j <= len2; j ++)
        {
            for(k = 0; k <= len; k ++)
            {
                if(maxz < dp[i][j][k])
                {
                    maxz = dp[i][j][k];
                    a = i;
                    b = j;
                    kk = k;
                }
            }
        }
    }
    if(maxz == 0)
    {
        printf("0
");
        return 0;
    }
    int num = 0;
    //printf("%d
",maxz);
    while(a != 0&&b != 0)
    {
        if(pre[a][b][kk] == 1)
        {
            ans[num++] = s1[a-1];
            kk = pre1[a][b][kk];
            a --;
            b --;
        }
        else if(pre[a][b][kk] == 2)
        {
            kk = pre1[a][b][kk];
            a --;
        }
        else if(pre[a][b][kk] == 3)
        {
            kk = pre1[a][b][kk];
            b --;
        }
        else
            break;
    }
    for(i = num-1; i >= 0; i --)
    {
        printf("%c",ans[i]);
    }
    printf("
");
    return 0;
}