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  • bzoj1030 AC自动机+dp

    思路:建状态图,在状态图上dp。

    #include<bits/stdc++.h>
    #define LL long long
    #define ll long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PII pair<int, int>
    #define y1 skldjfskldjg
    #define y2 skldfjsklejg
    
    using namespace std;
    
    const int N = 6000 + 7;
    const int M = 1e7 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 10007;
    
    int n, m;
    char s[N];
    
    void add(int &a, int b) {
        a += b; if(a >= mod) a -= mod;
    }
    struct Ac {
        int ch[N][26], dp[N][100][2], val[N], f[N], tot, sz;
        void init(int sz) {this->sz = sz, tot = 0;};
        int newNode() {
            tot++; f[tot] = 0; val[tot] = 0;
            memset(ch[tot], 0, sizeof(ch[tot]));
            return tot;
        }
        inline int idx(char c) {return c - 'A';}
        void addStr(char *s) {
            int u = 0;
            for(int i = 0; s[i]; i++) {
                int c = idx(s[i]);
                if(!ch[u][c]) ch[u][c] = newNode();
                u = ch[u][c];
            }
            val[u] = 1;
        }
        void build() {
            queue<int> que;
            for(int c = 0; c < sz; c++) {
                int v = ch[0][c];
                if(!v) ch[0][c] = 0;
                else f[v] = 0, que.push(v);
            }
            while(!que.empty()) {
                int u = que.front(); que.pop();
                val[u] |= val[f[u]];
                for(int c = 0; c < sz; c++) {
                    int v = ch[u][c];
                    if(!v) ch[u][c] = ch[f[u]][c];
                    else f[v] = ch[f[u]][c], que.push(v);
                }
            }
        }
    
        void solve() {
            memset(dp, 0, sizeof(dp));
            dp[0][0][0] = 1;
            for(int k = 0; k < m; k++) {
                for(int u = 0; u <= tot; u++) {
                    for(int c = 0; c < sz; c++) {
                        int v = ch[u][c];
                        add(dp[v][k + 1][1], dp[u][k][1]);
                        if(val[v]) add(dp[v][k + 1][1], dp[u][k][0]);
                        else add(dp[v][k + 1][0], dp[u][k][0]);
                    }
                }
            }
    
            int ans = 0;
            for(int i = 0; i <= tot; i++)
                add(ans, dp[i][m][1]);
            printf("%d
    ", ans);
        }
    } ac;
    int main() {
        ac.init(26);
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++) {
            scanf("%s", s);
            ac.addStr(s);
        }
        ac.build();
        ac.solve();
        return 0;
    }
    
    
    /*
    2
    abad
    ba
    -----
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/9491892.html
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