AC自动机专题:dp,巧妙的状态定义-HDU3341

题目大意:

给你一个只含"AGCT"四种字符的长度为 n($n\le 40$)的字符串S.然后给m($m\le 50$)个长度小于等于10的匹配串。你现在可以重新排列S使得这m个匹配串在其中出现的次数最多.输出这个次数.

题目思路:

建AC自动机后

其实不难想到一个朴素的转移方程:$dp(a_0,a_1,a_2,a_3,pos)$代表字符串S还剩下 $a_0$个A , $a_1$个G , $a_2$个C , $a_3$个T 且当前在Trie图中的节点为$pos$的出现最多次数。

但是问题是A,G,C,T都可能出现40次。这里我就蠢了。因为 A+G+C+T=40.所以根据均值不等式,状态数最多为: $10^4=10000$ 种。而不是$40^4$了. 这样我们就必须给状态编号.然后直接刷表法dp即可。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 500 + 5;
int ha[42][42][42][42] , num[10] , cnt , dp[14641][505] , bk[1005];
struct S
{
    
    
    int a[4];
}sta[14641];
namespace AC{
    
    
    int tr[maxn][5] , tot;
    int fail[maxn] , v[maxn];
    void init ()
    {
    
    
        tot = 0;
        for (int i = 0 ; i < maxn ; i++){
    
    
            for (int j = 0 ; j < 4 ; j++) tr[i][j] = 0;
            fail[i] = v[i] = 0;
        }
    }
    void add (char * s){
    
    
        int u = 0;
        for (int i = 1 ; s[i] ; i++){
    
    
            if (!tr[u][bk[s[i]]]) tr[u][bk[s[i]]] = ++tot;
            u = tr[u][bk[s[i]]];
        }
        v[u]++;
    }
    queue<int>q;
    void build (){
    
    
        while (q.size()) q.pop();
        for (int i = 0 ; i < 4 ; i++) if (tr[0][i]) q.push(tr[0][i]);
        while (q.size()){
    
    
            int u = q.front();q.pop();
            for (int i = 0 ; i < 4 ; i++){
    
    
                if (tr[u][i]){
    
    
                    fail[tr[u][i]] = tr[fail[u]][i];
                    q.push(tr[u][i]);
                }else {
    
    
                    tr[u][i] = tr[fail[u]][i];
                }
            }
            v[u] += v[fail[u]];
        }
    }
}
char tmp[15] , s[1005];
int main()
{
    
    
    int n;
    bk['A'] = 0;
    bk['G'] = 1;
    bk['C'] = 2;
    bk['T'] = 3;
    int t = 0;
    while ( scanf("%d" , &n) , n ){
    
    
        AC::init();
        for (int i = 1 ; i <= n ; i++){
    
    
            scanf("%s" , tmp + 1);
            AC::add(tmp);
        }
        AC::build();
        scanf("%s" , s + 1);
        memset(num , 0 , sizeof num);
        for (int i = 1 ; s[i] ; i++) num[bk[s[i]]]++;
        cnt = 0;
        for (int i = 0 ; i <= num[0] ; i++)
            for (int j = 0 ; j <= num[1] ; j++)
                for (int k = 0 ; k <= num[2] ; k++)
                    for (int l = 0 ; l <= num[3] ; l++){
    
    
                        ha[i][j][k][l] = cnt++ ,
                        sta[cnt - 1].a[0] = i , sta[cnt - 1].a[1] = j , sta[cnt - 1].a[2] = k , sta[cnt - 1].a[3] = l;
                        for (int gg = 0 ; gg <= AC::tot ; gg++) dp[cnt - 1][gg] = -1;
                    }
        dp[ha[num[0]][num[1]][num[2]][num[3]]][0] = 0;
        for (int i = cnt - 1 ; i > 0 ; i--){
    
    
            for (int j = 0 ; j <= AC::tot ; j++){
    
    
                if (dp[i][j] == -1) continue;
                for (int k = 0 ; k < 4 ; k++){
    
    
                    if (sta[i].a[k] == 0) continue;
                    sta[i].a[k]--;
                    int ni = ha[sta[i].a[0]][sta[i].a[1]][sta[i].a[2]][sta[i].a[3]] , nj = AC::tr[j][k];
                    dp[ni][nj] = max (dp[ni][nj] , dp[i][j] + AC::v[nj]);
                    sta[i].a[k]++;
                }
            }
        }
        int ans = 0;
        for (int i = 0 ; i <= AC::tot ; i++) if (dp[0][i] != -1) ans = max (ans , dp[0][i]);
        printf("Case %d: %d\n" , ++t , ans);
    }
    return 0;
}