HDU-2457 DNA Repair

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 retried red 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 retried red.

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
Number of characters which need to be changed. If it's impossible to repair the given DNA, print-1.

Sample Input

 2AAAAAGAAAG    2ATGTGAATG4AGCTAGT0 

 

Sample output

 Case 1: 1Case 2: 4Case 3: -1 

Question: give some invalid pattern DNA strings, give an original string, and ask how many characters need to be modified at least so that the original string does not contain illegal string ideas: first, we construct an AC automatic machine. To ensure that the child strings do not match, we stipulate that we do not go to the end of each sub-string, and then we are in the DP except the inaccessible location, if DP [I] [J] is set, it indicates that the parent string goes to the I-th word, and the automatic machine matches the minimum value of the J state, when we encounter a different letter, we need to modify the substring so that it is the same as the mother string. If it is the same, we will not change it. Now we try to change it to the mother string.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int inf = 0x3f3f3f3f;struct Trie {    int next[1010][4], fail[1010];    int end[1010];    int root, L;    int newNode() {        for (int i = 0; i < 4; i++)            next[L][i] = -1;        end[L++] = false;        return L-1;    }    void init() {        L = 0;        root = newNode();    }    int getInt(char ch) {        if (ch == ‘A‘) return 0;        else if (ch == ‘C‘) return 1;        else if (ch == ‘G‘) return 2;        else if (ch == ‘T‘) return 3;    }    void insert(char buf[]) {        int len = strlen(buf);                int now = root;        for (int i = 0; i < len; i++) {            if (next[now][getInt(buf[i])] == -1)                next[now][getInt(buf[i])] = newNode();            now = next[now][getInt(buf[i])];        }        end[now] = 1;    }    void build() {        queue<int> Q;        fail[root] = root;        for (int i = 0; i < 4; i++) {            if (next[root][i] == -1)                 next[root][i] = root;            else {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        }        while (!Q.empty()) {            int now = Q.front();            Q.pop();            if (end[fail[now]]) end[now] = 1; //notice            for (int i = 0; i < 4; i++)                 if (next[now][i] == -1)                    next[now][i] = next[fail[now]][i];                else {                    fail[next[now][i]] = next[fail[now]][i];                    Q.push(next[now][i]);                }        }    }    int dp[1010][1010];    int solve(char buf[]) {        int len = strlen(buf);        for (int i = 0; i <= len; i++)            for (int j = 0; j < L; j++)                dp[i][j] = inf;        dp[0][root] = 0;        for (int i = 0; i < len; i++)             for (int j = 0; j < L; j++)                 if (dp[i][j] != inf) {                    for (int k = 0; k < 4; k++) {                        int news = next[j][k];                        if (end[news]) continue;                        int tmp;                        if (k == getInt(buf[i]))                            tmp = dp[i][j];                        else tmp = dp[i][j] + 1;                        dp[i+1][news] = min(dp[i+1][news], tmp);                    }                }        int ans = inf;        for (int j = 0; j < L; j++)            ans = min(ans, dp[len][j]);        if (ans == inf)             ans = -1;        return ans;    }} ac;char buf[1010];int main() {    int n, cas = 1;    while (scanf("%d", &n) != EOF && n) {        ac.init();        while (n--) {            scanf("%s", buf);            ac.insert(buf);        }        ac.build();        scanf("%s", buf);        printf("Case %d: %d\n", cas++, ac.solve(buf));    }    return 0;}

HDU-2457 DNA Repair