POJ 3691 DNA repair【AC自動機+DP】

只能說這道題目的數據好強啊，以前寫的自動機都是有缺陷的但是可以過題，這次被坑大發了……嗚嗚

f(i, j)表示長度爲i的串，到j狀態時需要更改的次數。

f(i, u)  =  min(f(i, u), f(i-1, j) + (s[i] != k))

u是狀態j的子狀態。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
struct AC_Automata {
    #define N 1003
    #define M 4
    int ch[N][M], val[N], last[N], f[N], sz;
    void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }
    int idx(char c) {
        if (c == 'A') return 0;
        if (c == 'G') return 1;
        if (c == 'C') return 2;
        return 3;
    }

    void insert(char s[], int v) {
        int u = 0;
        for (int i=0; s[i]; i++) {
            int c = idx(s[i]);
            if (!ch[u][c]) {
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] = 1;
    }
    void build() {
        queue<int> q;
        f[0] = 0;
        for (int c=0; c<M; c++) {
            int u = ch[0][c];
            if (u) { f[u] = last[u] = 0; q.push(u); }
        }
        while (!q.empty()) {
            int r = q.front(); q.pop();
            for (int c=0; c<M; c++) {
                int u = ch[r][c];
                val[r] = val[r] || val[f[r]]; ///傳遞後綴串的合法性
                if (!u) { ch[r][c] = ch[f[r]][c]; continue; }
                q.push(u);
                f[u] = ch[f[r]][c];
                last[u] = val[f[u]] ? f[u] : last[f[u]];
            }
        }
    }
} ac;
int f[1003][1003];
#define inf 0x3f3f3f3f
char s[1005];
void solve(int T) {
    scanf(" %s", s);
    int len = strlen(s);
    memset(f, 0x3f, sizeof(f));
    f[0][0] = 0;
    int u, t;
    for (int i=1; i<=len; i++) {
        for (int j=0; j<ac.sz; j++) {
            if (f[i-1][j] >= inf) continue;
            for (int k=0; k<4; k++) {
                u = ac.ch[j][k];
                if (ac.val[u]) continue;
                if (ac.idx(s[i-1]) == k) t = 0;
                else t = 1;
                f[i][u] = min(f[i][u], f[i-1][j]+t);
            }
        }
    }
    int ans = inf;
    for (int i=0; i<ac.sz; i++)
        if (ac.val[i] == 0) ans = min(ans, f[len][i]);

    printf("Case %d: ", T);
    if (ans == inf) puts("-1");
    else printf("%d\n", ans);
}
int main() {
    int n, T = 0;

    while (scanf("%d", &n) == 1 && n) {
        ac.clear();
        for (int i=1; i<=n; i++) {
            scanf(" %s", s); ac.insert(s, 1);
        }
        ac.build();

        solve(++T);
    }
    return 0;
}