HDU 2457 DNA_repair

DNA repair

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2530    Accepted Submission(s): 1356
Problem Description

Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a DNA is represented as a string containing characters 'A', 'G' , 'C' and 'T'. The repairing techniques are simply to change some characters to eliminate all segments causing diseases. For example, we can repair a DNA "AAGCAG" to "AGGCAC" to eliminate the initial causing disease segments "AAG", "AGC" and "CAG" by changing two characters. Note that the repaired DNA can still contain only characters 'A', 'G', 'C' and 'T'.

You are to help the biologists to repair a DNA by changing least number of characters.

 

Input

The input consists of multiple test cases. Each test case starts with a line containing one integers N (1 ≤ N ≤ 50), which is the number of DNA segments causing inherited diseases.
The following N lines gives N non-empty strings of length not greater than 20 containing only characters in "AGCT", which are the DNA segments causing inherited disease.
The last line of the test case is a non-empty string of length not greater than 1000 containing only characters in "AGCT", which is the DNA to be repaired.

The last test case is followed by a line containing one zeros.

 

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the
number of characters which need to be changed. If it's impossible to repair the given DNA, print -1.

 

Sample Input

2

AAA

AAG

AAAG

2

A

TG

TGAATG

4

A

G

C

T

AGT

0

 

Sample Output

Case 1: 1

Case 2: 4

Case 3: -1

 

基础的AC自动机DP，按长度一层层前向转移

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <vector>
#include <queue>
#include <time.h>
#define eps 1e-7
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define rep0(j,n) for(int j=0;j<n;++j)
#define rep1(j,n) for(int j=1;j<=n;++j)
#define pb push_back
#define set0(n) memset(n,0,sizeof(n))
#define ll long long
#define ull unsigned long long
#define iter(i,v) for(edge *i=head[v];i;i=i->next)
//#define max(a,b) (a>b?a:b)
//#define min(a,b) (a<b?a:b)
//#define OJ
using namespace std;
const int MAXINT = 1110;
const int MAXNODE = 1110;
const int MAXEDGE = 1110;
int n, T;
char s[MAXINT];
struct node {
    node *fail;
    node *c[4];
    int bad;
    int nd[MAXINT]; //for a string of length i, least change required to reach this state
};
int toint(char c) {
    switch (c) {
    case 'A': return 0;
    case 'C': return 1;
    case 'G': return 2;
    case 'T': return 3;
    }
    return 0;
}
struct ac_automaton {
    node mp[MAXNODE];
    node *root, *pre;
    node *q[MAXNODE];
    int cnt;
    ac_automaton() {
        memset(mp, 0, sizeof(mp));
        memset(q, 0, sizeof(q));
        cnt = 0;
        root = new_node();
        pre = root;
    }
    void clear() {
        memset(mp, 0, sizeof(mp));
        memset(q, 0, sizeof(q));
        cnt = 0;
        root = new_node();
        pre = root;
    }
    node *new_node() {
        memset(mp[cnt].nd, 0x3f3f, sizeof(mp[cnt].nd));
        return &mp[cnt++];
    }
    void add(int w) {
        if (!pre->c[w]) pre->c[w] = new_node();
        pre = pre->c[w];
    }
    void danger() {
        pre->bad = 1;
    }
    void init() {
        pre = root;
    }
    void get_fail() {
        int h = 0, t = 0;
        node *p, *tmp;
        q[t++] = root;
        root->fail = NULL;
        while (h != t) {
            p = q[h++];
            rep0(i, 4) {
                if (p->c[i]) {
                    if (p == root) p->c[i]->fail = root;
                    else {
                        tmp = p->fail;
                        while (tmp) {
                            if (tmp->c[i]) {
                                p->c[i]->bad |= tmp->c[i]->bad;
                                p->c[i]->fail = tmp->c[i];
                                break;
                            }
                            tmp = tmp->fail;
                        }
                        if (!tmp) p->c[i]->fail = root;
                    }
                    q[t++] = p->c[i];
                }
                else {
                    p->c[i] = p->fail&&p->fail->c[i] ? p->fail->c[i] : root;
                }
            }
        }
    }
    void solve(char *s) {
        int l = strlen(s);
        root->nd[0] = 0;
        rep0(i, l) {
            rep0(j, cnt) {
                if (mp[j].nd[i] != INF) {
                    rep0(k, 4) {
                        node *p = mp[j].c[k];
                        if (!p || p->bad) continue;
                        p->nd[i + 1] = min(p->nd[i + 1], mp[j].nd[i] + ((toint(s[i]) == k) ? 0 : 1));
                    }
                }
            }
        }
        int ans = INF;
        rep0(i, cnt) {
            if (!mp[i].bad) {
                ans = min(ans, mp[i].nd[l]);
            }
        }
        printf("Case %d: %d
", T, ans == INF ? -1 : ans);
    }
} acm;
int read() {
    char c = getchar(); int f = 1, x = 0;
    while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
    while (isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return f * x;
}
char get_ch() {
    char c = getchar();
    while (!isalpha(c)) c = getchar();
    return c;
}

void get_input();
void work();
int main() {
    freopen("in.txt", "r", stdin);
    freopen("out.wa", "w", stdout);
    while (++T) {
        scanf("%d", &n);
        if (!n) break;
        acm.clear();
        get_input();
        work();
    }
    return 0;
}
void work() {
    acm.solve(s);
}
void get_input() {
    rep0(i, n) {
        scanf("%s", s);
        int l = strlen(s);
        rep0(j, l) acm.add(toint(s[j]));
        acm.danger();
        acm.init();
    }
    acm.get_fail();
    scanf("%s", s);
}