[回文自动机][DP] Codeforces gym 100543 G - Virus synthesis

最新推荐文章于 2020-11-16 22:12:27 发布

Vectorxj

最新推荐文章于 2020-11-16 22:12:27 发布

阅读量357

点赞数 1

CC 4.0 BY-SA版权

分类专栏：回文自动机动态规划

本文链接：https://blog.youkuaiyun.com/Vectorxj/article/details/79516683

动态规划同时被 2 个专栏收录

62 篇文章

订阅专栏

回文自动机

4 篇文章

订阅专栏

本文介绍了一种基于回文自动机的动态规划算法，用于计算将任意字符串转换为回文串所需的最少操作次数。该算法通过维护每个节点代表的回文串长度及对应的最少操作次数来实现高效求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

$Solution$

考虑在回文自动机上DP
$f_u$ 表示回文自动机中 $u$ 节点所代表的回文串最少需要几次才能得到

l e n_{u}

$len_u$ 为奇数，

fu=lenufu=lenu $f_u=len_u$ 。

$len_u$ 为偶数， $f_u=\text{min}\big(f_p+1,f_{half_u}+({len_u\over2}-len_{half_u})+1\big)$

#include <bits/stdc++.h>
using namespace std;

const int N = 233333;

int test, n;
char chr[N];
int s[N];
int ch[N][10];
int par[N], len[N];
int rgt[N], half[N], f[N];
int last, tcnt;

inline void init(void) {
    for (int i = 0; i <= tcnt; i++) {
        half[i] = 0;
        for (int j = 0; j < 5; j++)
            ch[i][j] = 0;
    }
    par[0] = par[1] = 1;
    half[0] = half[1] = 1;
    len[1] = -1;
    f[0] = 1; f[1] = -1;
    tcnt = 1; last = 0;
}
inline void extend(int pos) {
    int p = last, key = s[pos];
    while (s[pos - len[p] - 1] != key) p = par[p];
    if (!ch[p][key]) {
        int np = ++tcnt, q = par[p];
        len[np] = len[p] + 2;
        while (s[pos - len[q] - 1] != key) q = par[q];
        par[np] = ch[q][key];
        q = half[p]; 
        while (len[ch[q][key]] > len[np] / 2 || s[pos - len[q] - 1] != key) q = par[q];
        half[np] = ch[q][key];
        if (len[np] & 1) {
            f[np] = len[np];
        } else {
            f[np] = min(f[p] + 1, f[half[np]] + len[np] / 2 - len[half[np]] + 1);
        }
        ch[p][key] = np;
    }
    last = ch[p][key];
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    scanf("%d\n", &test);
    while (test--) {
        scanf("%s", chr + 1);
        n = strlen(chr + 1);
        for (int i = 1; i <= n; i++)
            if (chr[i] == 'A') s[i] = 1;
            else if (chr[i] == 'T') s[i] = 2;
            else if (chr[i] == 'G') s[i] = 3;
            else if (chr[i] == 'C') s[i] = 4;
        init(); int ans = 1 << 30;
        for (int i = 1; i <= n; i++) {
            extend(i);
            ans = min(f[last] + n - len[last], ans);
        }
        cout << ans << endl;
    }
    return 0;
}