poj3267The Cow Lexicon(dp)
http://poj.org/problem?id=3267
对dp没研究 看了看解题报告 还是比较好推的 dp[i] = min{dp[i]-1,dp[j]+i-j-length[k]} 若从J到i包含字典中的单词
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 char word[650][30]; 7 int main() 8 { 9 int i,j,k,n,m,dp[310],kk[610]; 10 char str[310]; 11 while(cin>>n>>m) 12 { 13 cin>>str; 14 memset(dp,0,sizeof(dp)); 15 for(i = 1; i <= n; i++) 16 { 17 cin>>word[i]; 18 kk[i] = strlen(word[i]); 19 } 20 for(i = 0 ; i < m ; i++) 21 { 22 int mi = 1000; 23 if(i>0) 24 dp[i] = dp[i-1]+1; 25 else 26 dp[i] = 1; 27 for(j = 1 ; j <= n ; j++) 28 { 29 int g = kk[j]-1,tk = i; 30 while(tk>=0&&g>=0&&(i-tk+1-kk[j])<dp[i]) 31 { 32 if(str[tk]==word[j][g]) 33 g--; 34 tk--; 35 } 36 if(tk<0) 37 tk = 0; 38 if(g<0) 39 { 40 mi = min(mi,i-tk-kk[j]+dp[tk]); 41 } 42 else 43 mi = min(mi,i-tk+dp[tk]); 44 } 45 if(mi!=1000) 46 dp[i] = min(dp[i],mi); 47 } 48 cout<<dp[m-1]<<endl; 49 } 50 return 0; 51 }