UVA - 11552 Fewest Flops (dp)

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);
    }
}
/*
2
5 helloworld
7 thefewestflops
*/