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  • AC自动机+DP HDOJ 2457 DNA repair(DNA修复)

    题目链接

    题意:

      给n串有疾病的DNA序列,现有一串DNA序列,问最少修改几个DNA,能使新的DNA序列不含有疾病的DNA序列。

    思路:

      构建AC自动机,设定end结点,dp[i][j]表示长度i的前缀串走到自动机的j点最少需要修改几个DNA。状态转移方程。那么只要转移到下一个的DNA不是end结点就能转移,如果下一个DNA不和原序列不一样就+1。

    #include <bits/stdc++.h>
    
    const int N = 50 + 5;
    const int M = 1000 + 5;
    const int INF = 0x3f3f3f3f;
    struct AC {
        static const int NODE = N * 20;
        static const int SIZE = 4;
        int ch[NODE][SIZE], fail[NODE], end[NODE];
        int sz;
    
        void init();
        int idx(char c);
        void insert(char *s);
        void get_fail();
    }ac;
    
    char s[25];
    char str[M];
    int dp[M][AC::NODE];
    
    int solve() {
        int len = strlen (str);
        memset (dp, INF, sizeof (dp));
        dp[0][0] = 0;
    
        for (int i=1; i<=len; ++i) {
            for (int j=0; j<ac.sz; ++j) {
                if (dp[i-1][j] == INF) continue;
                for (int k=0; k<4; ++k) {
                    if (!ac.end[ac.ch[j][k]]) {
                        dp[i][ac.ch[j][k]] = std::min (dp[i][ac.ch[j][k]], dp[i-1][j] + (ac.idx (str[i-1]) != k));
                    }
                }
            }
        }
    
        int ret = INF;
        for (int i=0; i<ac.sz; ++i) {
            if (!ac.end[i]) {
                ret = std::min (ret, dp[len][i]);
            }
        }
        if (ret == INF) ret = -1;
        return ret;
    }
    
    int main() {
        int n, cas = 0;
        while (scanf ("%d", &n) == 1 && n) {
            ac.init ();
            for (int i=0; i<n; ++i) {
                scanf ("%s", s);
                ac.insert (s);
            }
            ac.get_fail ();
            scanf ("%s", str);
    
            printf ("Case %d: %d
    ", ++cas, solve ());
        }
        return 0;
    }
    
    void AC::init() {
        memset (ch[0], 0, sizeof (ch[0]));
        sz = 1;
    }
    
    int AC::idx(char c) {
        if (c == 'A') {
            return 0;
        } else if (c == 'G') {
            return 1;
        } else if (c == 'C') {
            return 2;
        } else {
            return 3;
        }
    }
    
    void AC::insert(char *s) {
        int u = 0;
        for (int c, i=0; s[i]; ++i) {
            c = idx (s[i]);
            if (!ch[u][c]) {
                memset (ch[sz], 0, sizeof (ch[sz]));
                end[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        end[u] = 1;
    }
    
    void AC::get_fail() {
        fail[0] = 0;
        std::queue<int> que;
        for (int c=0; c<SIZE; ++c) {
            int u = ch[0][c];
            if (u) {
                fail[u] = 0;
                //last[u] = 0;
                que.push (u);
            }
        }
        while (!que.empty ()) {
            int r = que.front ();
            que.pop ();
            for (int c=0; c<SIZE; ++c) {
                int &u = ch[r][c];
                if (!u) {
                    u = ch[fail[r]][c];
                } else {
                    int v = fail[r];
                    while (v && !ch[v][c]) v = fail[v];
                    fail[u] = ch[v][c];
                    end[u] |= end[fail[u]];
                    //last[u] = end[fail[u]] ? fail[u] : last[fail[u]];
                    que.push (u);
                }
            }
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/5669431.html
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