HDU 2457 POJ 3691 DNA repair（AC自动机+DP）

给(n)个模式串和一个串(s)，要求修改(s)最少的字符使得没有一个模式串是(s)的子串

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(dp[i][j])表示长度为(i)，到达(AC)自动机的节点(j)所需修改的最少字符数量

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2000;
struct Trie{
    int next[maxn][26], fail[maxn], end[maxn];
    int dp[maxn][maxn];
    int last[maxn];
    int root, L;
    int newnode(){
        for (int i = 0; i < 26; i++)
            next[L][i] = -1;
        fail[L]=last[L]=end[L]=0;
        return L++;
    }
    void init(){
        L = 0;
        root = newnode();
    }
    void insert(char buf[]){
        int len = strlen(buf);
        int now = root;
        for(int i=0;i<len;i++){
            if(next[now][buf[i]-'A'] == -1) 
                next[now][buf[i]-'A'] = newnode();
            now = next[now][buf[i]-'A'];
        }
        end[now]++;
    }
    void build(){
        queue<int>Q;
        fail[root]=root;
        for(int i = 0;i < 26;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();
            end[now]+=end[fail[now]];
            for(int i = 0;i < 26;i++){
                if(next[now][i] == -1) 
                    next[now][i] = next[fail[now]][i];
                else{
                    int son=next[now][i];
                    fail[son]=next[fail[now]][i];
                    last[son]=end[fail[son]]?fail[son]:last[fail[son]];
                    Q.push(next[now][i]); 
                }  
            }
        }
    }
    char to[4]={'A','G','C','T'};
    void solve(char s[]){
        memset(dp,0x3f,sizeof(dp));
        dp[0][0]=0;
        int n=strlen(s);
        for(int i=1;i<=n;i++){
            for(int j=0;j<L;j++){
                if(end[j]||dp[i-1][j]==0x3f3f3f3f)continue;
                for(char k:to){
                    int son=next[j][k-'A'];
                    if(!end[son])
                        dp[i][son]=min(dp[i][son],dp[i-1][j]+(s[i-1]!=k));
                }
            }
        }
        int ans=0x3f3f3f3f;
        for(int i=0;i<L;i++)
            ans=min(ans,dp[n][i]);
        if(ans==0x3f3f3f3f)ans=-1;
        printf("%d
",ans);
    }
}ac;

char s[maxn];
int main(){
    int n,_=0;
    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: ",++_);
        ac.solve(s);
    }
    return 0;
}