洛谷 P2031 脑力达人之分割字串

题目传送门

解题思路:

f[i]表示到第i位可获得的最大分割次数,对于每个f[i]都可由其符合条件的前缀转移过来,条件就是当前串除了前缀的剩余字符里有所给单词,然后一看,这不是在剩余字符里找有没有所给单词吗?所以果断KMP,其实本题好像不用KMP,暴力模拟就可以,但是为了练习KMP装逼,所以就写一下.

AC代码:

 #include<iostream>
 #include<cstdio>

 using namespace std;

 string l,p,a[];
 int n,kmp[][],f[]; 

 inline int max(int s,int d) {
     if(s > d) return s;
     return d;
 }

 int main() {
     cin >> p;
     l = " " + p;
     scanf("%d",&n);
     for(int i = ;i <= n; i++) {
         cin >> p;
         a[i] = " " + p;
         int len = a[i].length() - ,k = ;
         for(int j = ;j <= len; j++) {
             while(k !=  && a[i][j] != a[i][k+]) k = kmp[i][k];
             if(a[i][j] == a[i][k+]) k++;
             kmp[i][j] = k;
         }
     }
     for(int i = ;i <= l.length() - ; i++) {
         for(int j = ;j <= i; j++) {
             string u = " " + l.substr(j,i-j+);
             int o;
             bool vis = ;
             for(int k = ;k <= n; k++) {
                 o = ;
                 for(int p = ;p <= u.length() - ; p++) {
                     while(o !=  && u[p] != a[k][o+]) o = kmp[k][o];
                     if(u[p] == a[k][o+]) o++;
                     if(o == a[k].length() - ) {
                         vis = ;
                         break;
                     }
                 }
                 if(vis) break;
             }
             if(vis)
                 f[i] = max(f[i],f[j-] + );
         }
     }
     printf("%d",f[l.length()-]);
     return ;
 }