1 # include<cstdio>
 2 # include<cstring>
 3 # include<algorithm>
 4
 5 using namespace std;
 6
 7 const int maxn = 16 + 1;
 8 const int INF = 0xffffff;
 9 char s[maxn];
10 bool ispal[1<<maxn];    //ispal[i]表示状态i是否是回文，2进制时1表示选择这个字符，0表示不选择这个字符
11 int d[1<<maxn];        //d[S]表示将集合S中的字母删除需要多少步，结果就是d[(1<<n)-1]
12 int n;
13
14 void getPal()       //求出所有的回文串
15 {
16     int S, i, j;
17     for(S = 0; S < (1<<n); S++){    //状态S是否回文
18         bool ok = 1;
19         int m = 0, buf[maxn];    //临时存储提取出来的字符
20         for(i = 0; i < n; i++) if((1<<i) & S){    //提取对应字符
21             buf[m++] = s[i];
22         }
23         for(i = 0, j = m-1; i < j; i++, j--){ //判断回文
24             if(buf[i] != buf[j]){
25                 ok = 0;
26                 break;
27             }
28         }
29         ispal[S] = ok;
30     }
31 }
32
33 void dp(){
34     int S, sub;
35     d[0] = 0;
36     for(S = 1; S < (1<<n); S++){
37         d[S] = INF;
38         for(sub = S; sub > 0; sub = (sub-1) & S){    //sub = (sub-1) & S)保证是集合S中的字母
39             if(ispal[sub]) d[S] = min(d[S], d[S^sub] + 1);    //S^sub就是剩下的字符
40         }
41     }
42 }
43
44 int main()
45 {
46     int T;
47     scanf("%d", &T);
48     while(T--){
49         scanf("%s", s);
50         n = strlen(s);
51         getPal();
52         dp();
53         printf("%d\n", d[(1<<n)-1]);
54     }
55     return 0;
56 }

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 int min(int a,int b){
 7     return a<b ?a :b;
 8 }
 9
10 const int MAX_N = 16, INF = 0xffffff;
11 int n;
12 int dp[1 << MAX_N][MAX_N][MAX_N]; //rest,i,j
13 char s[MAX_N + 1];
14 void work() {
15     int i,j,k;
16     scanf("%s", s);
17     n = strlen(s);
18
19     for (i = 0; i < n; i++) {
20         for (j = i; j < n; j++) {
21             dp[0][i][j] = 0;
22         }
23     }
24
25     for (int rest = 1; rest < (1 << n); rest++) {
26         for (i = n - 1; i >= 0; i--) {
27             for (j = i; j < n; j++) {
28                 //rest,i,j
29                 int &ret = dp[rest][i][j] = INF;
30                 if (i < j)
31                     ret = min(dp[rest][i + 1][j], dp[rest][i][j - 1]);
32                 if (s[i] == s[j] && ((rest >> i) & 1) && ((rest >> j) & 1)) {
33                     int nrest = rest & (~(1 << i)) & (~(1 << j));
34                     if (nrest == 0)
35                         ret = min(ret, dp[nrest][i][j] + 1);
36                     else
37                         ret = min(ret, dp[nrest][i][j]);
38                 }
39             }
40         }
41         for (i = n - 1; i >= 0; i--) {
42             for (int j = i; j < n; j++) {
43                 dp[rest][i][j] = min(dp[rest][i][j], dp[rest][0][n - 1] + 1);
44             }
45         }
46     }
47
48     cout << dp[(1 << n) - 1][0][n - 1] << endl;
49 }
50
51 int main() {
52     int T;
53     cin >> T;
54     while (T--) {
55         work();
56     }
57 }