POJ3691 DNA repair(AC自动机 DP)

给定N个长度不超过20的模式串，再给定一个长度为M的目标串S，求在目标串S上最少改变多少字符，可以使得它不包含任何的模式串

建立Trie图，求得每个节点是否是不可被包含的串，然后进行DP

dp[i][j]表示在Trie图上行走i步到达节点j的要修改的最小次数，则

 

dp[i + 1][chi[j][k]] = min(dp[i + 1][chi[j][k]], dp[i][j] + (ACGT[k] != str[i]))

chi[j][k]表示节点j的第k个子节点

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

const int N = 2000, CH = 4;
const int INF = 0x3F3F3F3F;
const char *ACGT = "ACGT";
int Hash[128];

struct trie{
    trie *next[CH];
    trie *fail;
    int cnt, id;
}tree[N];

bool flag[1000];
int chi[1008][4];

class AC_Auto{

    public:

    int nxt;
    trie *root;

    AC_Auto(){
        root = &tree[0];
        nxt=0;
        memset(&tree[0], 0, sizeof(trie));
    }

    void insert(char *s){
        trie *p = root;
        for(int i = 0; s[i]; i++){
            int c = Hash[s[i]];
            if(!p -> next[c]){
                memset(&tree[++nxt], 0, sizeof(trie));
                p -> next[c] = &tree[nxt];
                p -> next[c] -> id = nxt;
            }
            p = p -> next[c];
        }
        p -> cnt = 1;
        flag[p->id] = 1;
    }

    void build(){
        queue<trie *> q;
        q.push(root);
        root -> fail = NULL;
        while(!q.empty()){
            trie *cur = q.front();
            q.pop();
            if(cur -> fail && flag[cur->fail ->id ]){
                flag[cur -> id] = 1;
            }
            for(int i = 0; i < CH; i++){
                trie *son = cur -> next[i];
                trie *tp = (cur == root)? root: cur -> fail->next[i];
                if(son == NULL){
                    cur -> next[i] = tp;
                }else{
                    son -> fail = tp;
                    q.push(son);
                }
                son = cur ->next[i];
                chi[cur -> id][i] = son -> id;
            }
        }
    }

};

char str[1008];
int dp[1008][1000];
int main(){
    Hash['A'] = 0;
    Hash['C'] = 1;
    Hash['G'] = 2;
    Hash['T'] = 3;

    int n;
    int t = 0;
    while(~scanf("%d", &n) && n){
        AC_Auto ac;
        memset(flag, 0, sizeof(flag));
        memset(dp, 0x3F, sizeof(dp));
        for(int i = 0; i < n; i++){
            scanf("%s", str);
            ac.insert(str);
        }

        ac.build();

        dp[0][0] = 0;
        scanf("%s", str);
        int num = ac.nxt + 1;

        int len = 0;
        for(int i = 0; str[i]; i++, len++){
            for(int j = 0; j < num; j++){
                for(int k = 0; k < CH; k++){
                    if(flag[chi[j][k]] == 0){
                        dp[i + 1][chi[j][k]] = min(dp[i + 1][chi[j][k]], dp[i][j] + (ACGT[k] != str[i]));
                    }
                }
            }
        }
        int ans = INF;
        for(int i = 0; i < num ; i++){
            ans = min(ans, dp[len][i]);
        }
        if(ans >= INF){
            ans = -1;
        }
        printf("Case %d: %d\n", ++t, ans);
    }
    return 0;
}