狀態轉移算是比較好推把，但是匹配字串很容易會超時，跟以前某個題目的用法很類似，從第i個字母往後找假如到第j個字母匹配了一個單詞,這樣dp[j] = dp[i-1]+j-i-l[k]+1，取小即可。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 int dp[501]; 8 char str[601][31],s1[601],l[601]; 9 int num,n,len; 10 int main() 11 { 12 int i,j,k,u; 13 scanf("%d%d",&n,&len); 14 scanf("%s",s1); 15 for(i = 1; i <= n; i ++) 16 { 17 scanf("%s",str[i]); 18 l[i] = strlen(str[i]); 19 } 20 for(i = 0; i <= len-1; i ++) 21 { 22 dp[i] = i+1; 23 } 24 for(i = 0; i <= len-1; i ++) 25 { 26 for(j = 1; j <= n; j ++) 27 { 28 if(s1[i] == str[j][0]) 29 { 30 num = 1; 31 if(l[j] == 1) 32 { 33 dp[i] = min(dp[i],dp[i-1]); 34 for(u = i+1;u <= len-1;u ++) 35 { 36 if(i-1 >= 0) 37 dp[u] = min(dp[u],dp[i-1]+u-i-l[j]+1); 38 else 39 dp[u] = min(dp[u],u-i-l[j]+1); 40 } 41 } 42 else 43 { 44 for(k = i+1; k <= len-1; k ++) 45 { 46 if(str[j][num] == s1[k]) 47 { 48 num ++; 49 } 50 if(num == l[j]) 51 { 52 for(u = k;u <= len-1;u ++) 53 { 54 if(i-1 >= 0) 55 dp[u] = min(dp[u],dp[i-1]+u-i-l[j]+1); 56 else 57 dp[u] = min(dp[u],u-i-l[j]+1); 58 } 59 break; 60 } 61 } 62 } 63 } 64 } 65 } 66 printf("%d\n",dp[len-1]); 67 return 0; 68 }