题目链接： https://www.nowcoder.com/acm/contest/79/C
思路是DP， dp[i][j][k] 表示到第i位为止变换次数为j次，且最后一个字母为k时的最大长度。(k==1时，整个串以b结尾，k==0时以a结尾)那么有
当前位置是a时
dp[i][j][0] = max(dp[i-1][j][0] + 1, dp[i-1][j-1][1]+1)
dp[i][j][1] = dp[i-1][j][1]
当前位置为b时有类似的状态转移方程，具体看代码。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 char s[100005]; 5 int a[100005]; 6 int dp[100005][11][2]; 7 int n, m; 8 9 int main() 10 { 11 scanf("%d%d", &n, &m); 12 scanf("%s", s); 13 int i = 0; 14 while ( s[i] && s[i] == 'b') 15 ++i; 16 if (i == n) 17 { 18 printf("0 "); 19 return 0; 20 } 21 /* 22 23 memset(dp, 0, sizeof(dp)); 24 dp[i][0][0] = 1; 25 for (++i; i < n; ++i) 26 { 27 dp[i][0][0] = dp[i-1][0][0] + (s[i]=='a'?1:0); 28 for (int j = 1; j <= m; ++j) 29 { 30 if (s[i] == 'a') 31 { 32 dp[i][j][0] = max(dp[i-1][j][0] + 1, dp[i-1][j-1][1] + 1); 33 dp[i][j][1] = dp[i-1][j][1]; 34 } 35 else 36 { 37 dp[i][j][1] = max(dp[i-1][j][1] + 1, dp[i-1][j-1][0] + 1); 38 dp[i][j][0] = dp[i-1][j][0]; 39 } 40 } 41 } 42 int ans = 0; 43 for (int i = 0; i < n; ++i) 44 { 45 for (int j = 0; j <=m; ++j) 46 { 47 ans = max(ans, dp[i][j][0]); 48 ans = max(ans, dp[i][j][1]); 49 } 50 } 51 52 53 */ 54 int len = 1; 55 char ch = 'a'; 56 int cnt = 0; 57 for (++i;i<n;++i) 58 { 59 if (s[i] == ch) 60 { 61 ++len; 62 } 63 else 64 { 65 a[cnt++] = len; 66 len = 1; 67 ch = s[i]; 68 } 69 } 70 a[cnt++] = len; 71 memset(dp, 0, sizeof(dp)); 72 dp[0][0][0] = a[0]; 73 int ans = a[0]; 74 for (int i = 1; i < cnt; ++i) 75 { 76 dp[i][0][0] = dp[i-1][0][0] + ((i&1)?0:a[i]); 77 // dp[i][0][1] = 0; 78 for (int j = 1; j <= m; ++j) 79 { 80 if (i&1) // 'b' 81 { 82 dp[i][j][1] = max(dp[i-1][j][1] + a[i], dp[i-1][j-1][0] + a[i]); 83 dp[i][j][0] = dp[i-1][j][0]; 84 } 85 else 86 { 87 dp[i][j][0] = max(dp[i-1][j][0] + a[i], dp[i-1][j-1][1] + a[i]); 88 dp[i][j][1] = dp[i-1][j][1]; 89 } 90 ans = max(ans, dp[i][j][0]); 91 ans = max(ans, dp[i][j][1]); 92 } 93 ans = max(ans, dp[i][0][0]); 94 ans = max(ans, dp[i][0][1]); 95 } 96 printf("%d ", ans); 97 return 0; 98 }