hdoj 3341 Lost’s revenge 【AC自动机 + 变进制状态压缩dp】

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 510
#define INF 0x3f3f3f3f
#define debug printf("1\n");
using namespace std;
int k = 1;
struct Trie
{
    int next[MAXN][4], fail[MAXN], word[MAXN];
    int L, root;
    int newnode()
    {
        for(int i = 0; i < 4; i++)
            next[L][i] = -1;
        word[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    int getval(char c)
    {
        if(c == 'A') return 0;
        if(c == 'C') return 1;
        if(c == 'G') return 2;
        if(c == 'T') return 3;
    }
    void Insert(char *s)
    {
        int now = root;
        for(int i = 0; s[i]; i++)
        {
            if(next[now][getval(s[i])] == -1)
                next[now][getval(s[i])] = newnode();
            now = next[now][getval(s[i])];
        }
        word[now]++;
    }
    void Build()
    {
        queue<int> Q;
        fail[root] = root;
        for(int i = 0; i < 4; i++)
        {
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            word[now] += word[fail[now]];
            for(int i = 0; i < 4; i++)
            {
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    int num[4], Hash[4];
    int dp[15000][MAXN];//i状态时在Trie的j节点最大的数论水平
    void solve(char *s)
    {
        memset(num, 0, sizeof(num));
        for(int i = 0; s[i]; i++)
            num[getval(s[i])]++;
        Hash[3] = 1;//T字符权值
        Hash[2] = num[3] + 1;//G字符权值
        Hash[1] = (num[2] + 1) * (num[3] + 1);//C字符权值
        Hash[0] = (num[1] + 1) * (num[2] + 1) * (num[3] + 1);//A字符权值
        memset(dp, -INF, sizeof(dp));
        dp[0][0] = 0;
        for(int A = 0; A <= num[0]; A++)
            for(int C = 0; C <= num[1]; C++)
                for(int G = 0; G <= num[2]; G++)
                    for(int T = 0; T <= num[3]; T++)
                    {
                        int State = A * Hash[0] + C * Hash[1] + G * Hash[2] + T * Hash[3];
                        for(int j = 0; j < L; j++)
                        {
                            if(dp[State][j] == -INF) continue;
                            for(int k = 0; k < 4; k++)
                            {
                                if(k == 0 && A == num[0]) continue;//注意临界条件的判断
                                if(k == 1 && C == num[1]) continue;
                                if(k == 2 && G == num[2]) continue;
                                if(k == 3 && T == num[3]) continue;
                                int nextnode = next[j][k];
                                int nextState = State + Hash[k];
                                dp[nextState][nextnode] = max(dp[nextState][nextnode], dp[State][j] + word[nextnode]);
                            }
                        }
                    }
        int SumState = num[0] * Hash[0] + num[1] * Hash[1] + num[2] * Hash[2] + num[3] * Hash[3];
        int ans = 0;
        for(int i = 0; i < L; i++)
            ans = max(ans, dp[SumState][i]);
        printf("Case %d: %d\n", k++, ans);
    }
};
Trie ac;
char str[50];
int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        ac.init();
        for(int i = 0; i < n; i++)
            scanf("%s", str), ac.Insert(str);
        ac.Build();
        scanf("%s", str);
        ac.solve(str);
    }
    return 0;
}