题目传送门
1 /*
2 题意：将子符串分成k组，每组的字符顺序任意，问改变后的字符串最少有多少块
3 三维DP：可以知道，每一组的最少块是确定的，问题就在于组与组之间可能会合并块，总块数会-1。
4 dp[i][j]表示第i组以第j个字符结尾的最少块数，状态转移方程：dp[i][j] = min (dp[i][j], dp[i-1][l] + chunk - 1);
5 意思就是枚举上一组的所有字符，当出现在i组并且不是放到末尾，那么能-1
6 */
7 /************************************************
8 * Author :Running_Time
9 * Created Time :2015-8-7 11:08:46
10 * File Name :UVA_11552.cpp
11 ************************************************/
12
13 #include <cstdio>
14 #include <algorithm>
15 #include <iostream>
16 #include <sstream>
17 #include <cstring>
18 #include <cmath>
19 #include <string>
20 #include <vector>
21 #include <queue>
22 #include <deque>
23 #include <stack>
24 #include <list>
25 #include <map>
26 #include <set>
27 #include <bitset>
28 #include <cstdlib>
29 #include <ctime>
30 using namespace std;
31
32 #define lson l, mid, rt << 1
33 #define rson mid + 1, r, rt << 1 | 1
34 typedef long long ll;
35 const int MAXN = 1e3 + 10;
36 const int INF = 0x3f3f3f3f;
37 const int MOD = 1e9 + 7;
38 char str[MAXN];
39 int dp[MAXN][MAXN];
40 bool vis[130];
41
42 int main(void) { //UVA 11552 Fewest Flop
43 int T; scanf ("%d", &T);
44 while (T--) {
45 int k; scanf ("%d%s", &k, str + 1);
46 int len = strlen (str + 1);
47 memset (dp, INF, sizeof (dp));
48 for (int i=1; i<=len/k; ++i) {
49 memset (vis, false, sizeof (vis));
50 for (int j=(i-1)*k+1; j<=i*k; ++j) {
51 vis[str[j]] = true;
52 }
53 int chunk = 0;
54 for (int j='a'; j<='z'; ++j) {
55 if (vis[j]) chunk++;
56 }
57 if (i == 1) {
58 for (int j=1; j<=k; ++j) {
59 dp[i][j] = chunk;
60 }
61 continue;
62 }
63 for (int j=1; j<=k; ++j) {
64 int last = (i - 1) * k + j;
65 for (int l=1; l<=k; ++l) {
66 int pre = (i - 2) * k + l;
67 if (vis[str[pre]] && (chunk == 1 || str[pre] != str[last])) {
68 dp[i][j] = min (dp[i][j], dp[i-1][l] + chunk - 1);
69 }
70 else dp[i][j] = min (dp[i][j], dp[i-1][l] + chunk);
71 }
72 }
73 }
74
75 int ans = INF;
76 for (int i=1; i<=k; ++i) {
77 ans = min (ans, dp[len/k][i]);
78 }
79 printf ("%d
", ans);
80 }
81
82 return 0;
83 }