题意：n个病毒串，给你一个串t，问你最少改几个能没有病毒串

思路：去年还觉得挺难得...其实就是AC自动机上跑一下简单的DP，每个位置都往没病毒的地方跑，然后看一下最少是什么。

代码：

#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1000 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 20090717;
int n, m;
int dp[maxn][maxn];
int getid(char s){
    if(s == 'A') return 0;
    if(s == 'C') return 1;
    if(s == 'G') return 2;
    return 3;
}
struct Aho{
    struct state{
        int next[4];
        int fail, cnt;
    }node[maxn];
    int size;
    queue<int> q;

    void init(){
        size = 0;
        newtrie();
        while(!q.empty()) q.pop();
    }

    int newtrie(){
        memset(node[size].next, 0, sizeof(node[size].next));
        node[size].cnt = node[size].fail = 0;
        return size++;
    }

    void insert(char *s){
        int len = strlen(s);
        int now = 0;
        for(int i = 0; i < len; i++){
            int c = getid(s[i]);
            if(node[now].next[c] == 0){
                node[now].next[c] = newtrie();
            }
            now = node[now].next[c];
        }
        node[now].cnt = 1;
    }

    void build(){
        node[0].fail = -1;
        q.push(0);

        while(!q.empty()){
            int u = q.front();
            q.pop();
            if(node[node[u].fail].cnt && u) node[u].cnt = 1;
            for(int i = 0; i < 4; i++){
                if(!node[u].next[i]){
                    if(u == 0)
                        node[u].next[i] = 0;
                    else
                        node[u].next[i] = node[node[u].fail].next[i];
                }
                else{
                    if(u == 0) node[node[u].next[i]].fail = 0;
                    else{
                        int v = node[u].fail;
                        while(v != -1){
                            if(node[v].next[i]){
                                node[node[u].next[i]].fail = node[v].next[i];
                                break;
                            }
                            v = node[v].fail;
                        }
                        if(v == -1) node[node[u].next[i]].fail = 0;
                    }
                    q.push(node[u].next[i]);
                }
            }
        }
    }

    void query(char *s){
        int len = strlen(s);

        for(int i = 0; i <= len; i++){
            for(int j = 0; j < size; j++){
                dp[i][j] = INF;
            }
        }
        dp[0][0] = 0;
        for(int i = 0; i < len; i++){
            for(int j = 0; j < size; j++){
                if(dp[i][j] == INF) continue;
                for(int k = 0; k < 4; k++){
                    if(node[node[j].next[k]].cnt) continue;
                    if(getid(s[i]) != k) dp[i + 1][node[j].next[k]] = min(dp[i + 1][node[j].next[k]], dp[i][j] + 1);
                    else dp[i + 1][node[j].next[k]] = min(dp[i + 1][node[j].next[k]], dp[i][j]);
                }
            }
        }
        int ans = INF;
        for(int i = 0; i < size; i++)
            if(node[i].cnt == 0) ans = min(ans, dp[len][i]);
        if(ans == INF) printf("-1\n");
        else printf("%d\n", ans);
    }

}ac;
char s[1005];
int main(){
    int ca = 1;
    while(~scanf("%d", &n) && n){
        ac.init();
        for(int i = 0; i < n; i++){
            scanf("%s", s);
            ac.insert(s);
        }
        ac.build();
        scanf("%s", s);
        printf("Case %d: ", ca++);
        ac.query(s);
    }
    return 0;
}