POJ1699 HDU 1560 Best Sequence(AC自动机 最短路)

曾写过迭代加深搜索的方法，现在使用在AC自动上跑最短路的方法

 

dp[i][j]表示状态为到节点i，模式串是否包含的状态为j的最短串的长度，则状态转移方程为:

 

dp[nx][ny] = min(dp[x][y] + 1) , 其中nx为x后继结点，ny为从y转移过来的新状态，更新时加入队列

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<string>
using namespace std;

int Hash[128];
typedef long long LL;

const int N = 150, CH = 4, INF = 0x3F3F3F3F;
int n, cnt;
struct Trie{
    Trie *next[CH];
    Trie *fail;
    int id;
}tree[N], *q[50000];

int chi[N][CH];
int sum[N];
int dp[N][1024];
bool inq[N][1024];

class ACauto{
    private:
        int nxt;
        Trie *root;

    public:
    ACauto(){
        root = &tree[0];
        nxt=0;
        memset(&tree[0], 0, sizeof(Trie));
    }

    void insert(string s, int id){
        Trie *p = root;
        for(int i = 0; i < s.size(); i++){
            int c = Hash[s[i]];
            if(!p -> next[c]){
                memset(&tree[++nxt], 0, sizeof(Trie));
                p -> next[c] = &tree[nxt];
                p -> next[c] -> id = nxt;
            }
            p = p -> next[c];
        }
        sum[p->id] = 1 <<id;
    }

    void build(){
        int front = 0, rear = 0;
        q[rear++] = root;
        root -> fail = NULL;
        while(front < rear){
            Trie *cur = q[front++];

            if(cur -> fail){
                sum[cur->id] |= sum[cur->fail ->id];
            }

            for(int i = 0; i < CH; i++){

                Trie *son = cur -> next[i];
                Trie *tp = (cur == root)? root: cur -> fail->next[i];
                if(son == NULL){
                    cur -> next[i] = tp;
                }else{
                    son -> fail = tp;
                    q[rear++] = son;
                }
                son = cur -> next[i];
                chi[cur->id][i] = son->id;
            }
        }
    }

    void solve(){
        memset(dp, 0x3F, sizeof(dp));
        memset(inq, 0, sizeof(inq));
        queue<int> q;
        dp[0][0] = 0;
        inq[0][0] = 1;
        int st = 1 << cnt;
        q.push(0);
        while(!q.empty()){
            int u = q.front();
            q.pop();
            int x = u / st;
            int y = u % st;
            inq[x][y] = 0;
            for(int i = 0 ; i < CH; i++){
                int nx  = chi[x][i], ny = y | sum[chi[x][i]];
                if(dp[nx][ny] > dp[x][y] + 1){
                    dp[nx][ny] = dp[x][y] + 1;
                    if(!inq[nx][ny]){
                        q.push(nx * st + ny);
                        inq[nx][ny] = 1;
                    }
                }
            }
        }
        int ans = INF;
        for(int i = 0; i <= nxt; i++){
            ans = min(ans, dp[i][st - 1]);
        }
        printf("%d\n", ans);

    }

};

char str[100];
int main(){
    Hash['A'] = 0;
    Hash['C'] = 1;
    Hash['G'] = 2;
    Hash['T'] = 3;
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        set<string> st;
        for(int i = 0; i < n; i++){
            string str;
            cin>>str;
            st.insert(str);
        }
        ACauto ac;
        memset(sum, 0, sizeof(sum));

        cnt = 0;
        for(set<string>::iterator it = st.begin(); it != st.end(); it++){
            ac.insert(*it, cnt++);
        }
        ac.build();
        ac.solve();
    }

    return 0;
}