POJ 3691 DNA repair (DP+字符串)

题意：给出nn(1≤n≤50，1≤n≤50) 个病毒DNA序列，长度均不超过20。现在给出一个长度不超过1000的字符串，求至少要更换多少个字符，

才能使这个字符串不包含这些DNA序列。

析：利用前缀来做好状态转移。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

string str;
vector<string> v, prx;
int dp[maxn][maxn], nxt[maxn][4];
bool nt[maxn];

const char *ACGT = "ACGT";

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1 && n){
    string s;
    v.clear();  prx.clear();
    for(int i = 0; i < n; ++i){
      cin >> s;
      v.push_back(s);
    }
    cin >> str;
    for(int i = 0; i < n; ++i)
      for(int j = 0; j <= v[i].size(); ++j)
        prx.push_back(v[i].substr(0, j));
    sort(prx.begin(), prx.end());
    prx.erase(unique(prx.begin(), prx.end()), prx.end());

    for(int i = 0; i < prx.size(); ++i){
      nt[i] = false;
      for(int j = 0; j < n && !nt[i]; ++j)
        nt[i] |= v[j].size() <= prx[i].size() && prx[i].substr(prx[i].size()-v[j].size()) == v[j];
      if(nt[i])  continue;
      for(int j = 0; j < 4; ++j){
        s = prx[i] + ACGT[j];
        int k;
        while(1){
          k = lower_bound(prx.begin(), prx.end(), s) - prx.begin();
          if(k < prx.size() && prx[k] == s)  break;
          s = s.substr(1);
        }
        nxt[i][j] = k;
      }
    }

    memset(dp, INF, sizeof dp);
    dp[0][0] = 1;
    for(int i = 1; i < prx.size(); ++i)  dp[0][i] = 0;
    for(int t = 0; t < str.size(); ++t){
      for(int i = 0; i < prx.size(); ++i){
        if(nt[i])  continue;
        for(int j = 0; j < 4; ++j){
          int k = nxt[i][j];
          dp[t+1][k] = min(dp[t+1][k], dp[t][i] + (str[t] != ACGT[j]));
        }
      }
    }

    int ans = INF;
    for(int i = 0; i < prx.size(); ++i)  if(!nt[i])
      ans = min(ans, dp[str.size()][i]);
    printf("Case %d: %d\n", ++kase, ans == INF ? -1 : ans);
  }
  return 0;
}