18.10.29 POJ 3691 DNA repair（AC自动机+dp）

  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, 0, sizeof(tree));
104         memset(f, 0, 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 }