UVA

思路；用dp[i][j]表示到第i组末尾为字母j块的最小值，那么dp[i][j] = min(dp[i][j], dp[i-1][l] + f[i] - 1), 第i组和第i-1组含有字母l，并且l和j不同，否则dp[i][j] = min(dp[i][j],dp[i-1][l] + f[i]).f[i]为每组的不同字母数。

#include<cstdio>#include<set>#include<algorithm>#include<cstring>#include<iostream>#include<map>#include<queue>#include<vector>#include<string>#include<sstream>#include<cmath>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 100 + 20;const double EPS = 1e-5;const int mod = 1e8 + 7;typedef unsigned long long ull;typedef long long ll;int dx[] = {0, 0, -1, 1, -1, -1, 1, 1};int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};int a, b;int gcd(int x, int y){    return !y ? x : gcd(y, x % y);}int k;char s[1010];int f[1010];int d[1010][27];int dp[1010][27];int main(){    int T;    scanf("%d", &T);    while(T--){        scanf("%d%s", &k, s);        int len = strlen(s);        memset(f, 0, sizeof f);        memset(dp, INF, sizeof dp);        memset(d, 0, sizeof d);        int t = 0;        int n = len / k;        for(int i = 1; i <= len / k; ++i){            int cnt = 0;            for(; t < i * k; ++t){                int c = s[t] - 'a';                if(d[i][c] == 0){                    cnt++;                    d[i][c]++;                }                else d[i][c]++;            }            f[i] = cnt;        }        for(int i = 0; i < 26; ++i) dp[0][i] = 0;        for(int i = 1; i <= n; ++i){            for(int j = 0; j < 26; ++j){                for(int l = 0; l < 26; ++l){                    if(d[i][l] && d[i - 1][l] && (l != j || l == j && f[i] == 1)){                        dp[i][j] = min(dp[i][j], dp[i - 1][l] + f[i] - 1);                    }                    else                        dp[i][j] = min(dp[i][j], dp[i - 1][l] + f[i]);                }            }        }        int ans = INF;        for(int i = 0; i < 26; ++i)            ans = min(ans ,dp[n][i]);        printf("%d\n", ans);    }}/*25 helloworld7 thefewestflops*/