UVa 11552 Fewest Flops

dp[m][i][j]代表第m组以字母(i +'a') 开头和以字母 ( j + 'a' )结尾的最小块数。kuai[m]中存储了第 m 组最少可以分为多少块，显然最少为所有相同字母放在一起时的块数，即这一组中不同字母的个数。

如果前一组的结尾与后一组的开头相同，则dp[m][i][j] = min( dp[m-1][st][ed] + kuai[m] - 1 );

若不同，则dp[m][i][j] = min(dp[m - 1][st][ed] + kuai[m] );

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 1010;
const int INF = 1 << 30;

char str[MAXN];
int dp[MAXN][30][30];
int kuai[MAXN];       //记录每组之中最小的块数
bool vis[MAXN][30];   //记录每组出现了哪些字母
int K, n;

void GetKuai()     //预处理得到每组内最小的块数
{
    memset( vis, false, sizeof(vis) );
    for ( int i = 0; i < n; ++i )
    {
        int st = K * i;
        int ed = K * (i + 1);
        for ( int j = st; j < ed; ++j )
            vis[i][ str[j] - 'a' ] = true;

        kuai[i] = 0;
        for ( int j = 0; j < 26; ++j )
            if ( vis[i][j] ) ++kuai[i];
    }
    return;
}

void init()     //初始化
{
    for ( int i = 0; i < 26; ++i )
        for ( int j = 0; j < 26; ++j )
            if ( vis[0][i] && vis[0][j] )
                dp[0][i][j] = kuai[0];
            else dp[0][i][j] = INF;
    return;
}

void solved()
{
    init();

    for ( int m = 1; m < n; ++m )
        for ( int i = 0; i < 26; ++i )
        {
            if ( vis[m][i] )
            {
                for ( int j = 0; j < 26; ++j )
                {
                    dp[m][i][j] = INF;
                    if ( ( j == i && kuai[m] == 1 ) || ( j != i && vis[m][j] ) ) // 这里WA了两次，首位字母不能相同，除非这一组内的块数为1
                    {
                        for ( int st = 0; st < 26; ++st )
                            if ( vis[m - 1][st] )
                            {
                                for ( int ed = 0; ed < 26; ++ed )
                                {
                                    if ( ( st == ed && kuai[m - 1] == 1 ) || ( ed != st && vis[m - 1][ed] ) )
                                    {
                                        if ( ed == i )
                                            dp[m][i][j] = min( dp[m][i][j], dp[m - 1][st][ed] + kuai[m] - 1 );
                                        else dp[m][i][j] = min( dp[m][i][j], dp[m - 1][st][ed] + kuai[m] );
                                    }
                                }
                            }
                    }
                }
            }
        }

    int ans = INF;
    for ( int i = 0; i < 26; ++i )
        for ( int j = 0; j < 26; ++j )
            if ( vis[n - 1][i] && vis[n - 1][j] )
                ans = min( ans, dp[n - 1][i][j] );

    printf( "%d\n", ans );
    return;
}

int main()
{
    int T;
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d%s", &K, str );
        n = strlen(str) / K;   //得到块数
        GetKuai();
        solved();
    }
    return 0;
}