18.10.29 POJ 3691 DNA repairAC自动机+dp

描述

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 repaired DNA can still contain only characters ‘A‘,‘C‘ and ‘T‘.

You are to help the biologists to repair a DNA by changing least number of characters.

输入

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 repaired.

The last test case is followed by a line containing one zeros.

输出

For each test case,print a line containing the test case number( beginning with 1) followed by the
number of characters which need to be changed. If it‘s impossible to repair the given DNA,print -1.

 

样例输入

2
AAA
AAG
AAAG    
2
A
TG
TGAATG
4
A
G
C
T
AGT
0

样例输出

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

  1 #include <iostream>
  2 #include <string.h>
  3 #include <algorithm>
  4 #include <stack>
  5 #include <string>
  6 #include <math.h>
  7 #include <queue>
  8 #include <stdio.h>
  9 #include <string.h>
 10 #include <vector>
 11 #include <fstream>
 12 #include <set>
 13 #define inf 0x3f3f3f3f
 14 
 15 using namespace std;
 16 const int maxn = 1005;
 17 int n,nodecou,l;
 18 char str[maxn];
 19 int gen[85],f[maxn][maxn];
 20 
 21 struct node {
 22     int next[4];
 23     int prev;
 24     bool isdanger;
 25     char val;
 26     node() {
 27         memset(next,0,sizeof(next));
 28         prev = -1;
 29         isdanger = false;
 30         val = 0;
 31     }
 32 }tree[maxn];
 33 
 34 void insert() {
 35     int Now = 1,l=strlen(str);
 36     for (int i = 0; i < l; i++) {
 37         int idx = gen[str[i]];
 38         if (tree[Now].next[idx] == 0) {
 39             nodecou++;
 40             tree[Now].next[idx] = nodecou;
 41             tree[nodecou].val = str[i];
 42         }
 43         Now = tree[Now].next[idx];
 44         if (i == l - 1)
 45             tree[Now].isdanger = true;
 46     }
 47 }
 48 
 49 void build() {
 50     for(int i=0;i<4;i++)
 51         tree[0].next[i] = 1;
 52     tree[1].prev = 0;
 53     queue<int>q;
 54     q.push(1);
 55     while (!q.empty()) {
 56         int Now = q.front(); q.pop();
 57         for (int i = 0; i < 4; i++) {
 58             int child = tree[Now].next[i];
 59             int prev = tree[Now].prev;
 60             if (child) {
 61                 while (tree[prev].next[i] == NULL)
 62                     prev = tree[prev].prev;
 63                 tree[child].prev = tree[prev].next[i];
 64                 if (tree[tree[child].prev].isdanger)
 65                     tree[child].isdanger = true;
 66                 q.push(child);
 67             }
 68             else {
 69                 while (!tree[prev].next[i])
 70                     prev = tree[prev].prev;
 71                 tree[Now].next[i] = tree[prev].next[i];
 72                 if (tree[child].isdanger)
 73                     tree[Now].next[i] = 0;
 74             }
 75         }
 76     }
 77 }
 78 
 79 void dp(int kase) {
 80     for (int i = 1; i <= l; i++) {
 81         for (int j = 1; j <= nodecou; j++)
 82         {
 83             for (int k = 0; k < 4; k++)
 84                 if (tree[j].next[k] && !tree[tree[j].next[k]].isdanger)
 85                     f[i][tree[j].next[k]] = min(f[i - 1][j] + (gen[str[i-1]]!=k),f[i][tree[j].next[k]]);
 86         }
 87     }
 88     int ans = inf;
 89     for (int i = 1; i <= nodecou; i++)
 90         ans = min(ans,f[l][i]);
 91     if (ans < inf)
 92         printf("Case %d: %d\n",kase,ans);
 93     else
 94         printf("Case %d: -1\n",kase);
 95 }
 96 
 97 void init() {
 98     int kase = 0;
 99     while (scanf("%d",&n)) {
100         nodecou = 1;
101         if (n == 0)return;
102         kase++;
103         memset(tree,sizeof(tree));
104         memset(f,sizeof(f));
105         while (n--)
106         {
107             scanf("%s",str);
108             insert();
109         }
110         build();
111         scanf("%s",str);
112         l = strlen(str);
113         for (int i = 0; i <= l; i++)
114             for (int j = 1; j <= nodecou; j++)
115                 f[i][j] = inf;
116         f[0][1] = 0;
117         dp(kase);
118     }
119 }
120 
121 int main()
122 {
123     gen[‘A‘] = 0,gen[‘G‘] = 1,gen[‘C‘] = 2,gen[‘T‘] = 3;
124     init();
125     return 0;
126 }

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