DNA repair HDU - 2457 AC自动机+DP

题意：

给你N个模板串,并且给你一个文本串,

现在问你这个文本串最少需要改变几个字符才能使得它不包含任何模板串.

(以上字符只由A,T,G,C构成)

题解：

刚开始做这一题的时候表示很懵逼，好像没有学过这种类型的问题。

后面仔细想想，在之前的题目中，学会了求出不包含任何模板串的方案数。

这题可以转化下，求出所有不包含任何模板串的方案中与原串最少的不同数目。

根据这个DP

dp【i】【j】 表示走到长度为 i 的时候，在AC自动机 j 这个节点上最多与原串不同的个数。

然后这题就变成SB题了。

  1 #include <set>
  2 #include <map>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <unordered_map>
 14
 15 #define  pi    acos(-1.0)
 16 #define  eps   1e-9
 17 #define  fi    first
 18 #define  se    second
 19 #define  rtl   rt<<1
 20 #define  rtr   rt<<1|1
 21 #define  bug                printf("******\n")
 22 #define  mem(a, b)          memset(a,b,sizeof(a))
 23 #define  name2str(x)        #x
 24 #define  fuck(x)            cout<<#x" = "<<x<<endl
 25 #define  sfi(a)             scanf("%d", &a)
 26 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 27 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 28 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 29 #define  sfL(a)             scanf("%lld", &a)
 30 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 31 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 32 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 33 #define  sfs(a)             scanf("%s", a)
 34 #define  sffs(a, b)         scanf("%s %s", a, b)
 35 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 36 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 37 #define  FIN                freopen("../date.txt","r",stdin)
 38 #define  gcd(a, b)          __gcd(a,b)
 39 #define  lowbit(x)          x&-x
 40 #define  IO                 iOS::sync_with_stdio(false)
 41
 42
 43 using namespace std;
 44 typedef long long LL;
 45 typedef unsigned long long ULL;
 46 const ULL seed = 13331;
 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 48 const int maxn = 1e6 + 7;
 49 const int maxm = 8e6 + 10;
 50 const int INF = 0x3f3f3f3f;
 51 const int mod = 1e9 + 7;
 52
 53
 54 char str[55][22], buf[1010];
 55 int n, dp[1010][1010];
 56
 57 int get_num(char ch) {
 58     if (ch == ‘A‘) return 0;
 59     if (ch == ‘T‘) return 1;
 60     if (ch == ‘C‘) return 2;
 61     if (ch == ‘G‘) return 3;
 62 }
 63
 64 struct Aho_Corasick {
 65     int next[1010][4], fail[1010], End[1010];
 66     int root, cnt;
 67
 68     int newnode() {
 69         for (int i = 0; i < 4; i++) next[cnt][i] = -1;
 70         End[cnt++] = 0;
 71         return cnt - 1;
 72     }
 73
 74     void init() {
 75         cnt = 0;
 76         root = newnode();
 77     }
 78
 79     void insert(char buf[]) {
 80         int len = strlen(buf);
 81         int now = root;
 82         for (int i = 0; i < len; i++) {
 83             if (next[now][get_num(buf[i])] == -1) next[now][get_num(buf[i])] = newnode();
 84             now = next[now][get_num(buf[i])];
 85         }
 86         End[now]++;
 87     }
 88
 89     void build() {
 90         queue<int> Q;
 91         fail[root] = root;
 92         for (int i = 0; i < 4; i++)
 93             if (next[root][i] == -1) next[root][i] = root;
 94             else {
 95                 fail[next[root][i]] = root;
 96                 Q.push(next[root][i]);
 97             }
 98         while (!Q.empty()) {
 99             int now = Q.front();
100             Q.pop();
101              if (End[fail[now]]) End[now] = 1;
102             for (int i = 0; i < 4; i++)
103                 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
104                 else {
105                     fail[next[now][i]] = next[fail[now]][i];
106                     Q.push(next[now][i]);
107                 }
108         }
109     }
110
111     int solve() {
112         int len = strlen(buf);
113         for (int i = 0; i <= len; ++i)
114             for (int j = 0; j < cnt; ++j)
115                 dp[i][j] = INF;
116         dp[0][0] = 0;
117         for (int i = 0; i < len; ++i) {
118             for (int j = 0; j < cnt; ++j) {
119                 if (dp[i][j] == INF || End[j]) continue;
120                 for (int k = 0; k < 4; ++k) {
121                     int idx = next[j][k];
122                     if (End[idx]) continue;
123                     dp[i + 1][idx] = min(dp[i + 1][idx], dp[i][j] + (get_num(buf[i]) != k));
124                 }
125             }
126         }
127         int ans = INF;
128         for (int i = 0; i < cnt; ++i) ans = min(ans, dp[len][i]);
129         if (ans == INF) return -1;
130         return ans;
131     }
132
133     void debug() {
134         for (int i = 0; i < cnt; i++) {
135             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
136             for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
137             printf("]\n");
138         }
139     }
140 } ac;
141
142
143 int main() {
144     //FIN;
145     int cas = 1;
146     while (sfi(n) && n) {
147         ac.init();
148         for (int i = 1; i <= n; ++i) {
149             sfs(str[i]);
150             ac.insert(str[i]);
151         }
152         ac.build();
153         sfs(buf);
154         printf("Case %d: %d\n", cas++, ac.solve());
155     }
156     return 0;
157 }

原文地址：https://www.cnblogs.com/qldabiaoge/p/11379281.html