uva 11552 Fewest Flops 线性dp

// uva 11552 Fewest Flops//// 二维线性dp//// 首先，块内肯定是相同的字母放在一起，先记录下每个块内有多少种字母// 记作counts[i];// // 令f[i][j]表示前i个块以字母j为结尾的最小分块数//// 如果第i块的开始字母与第i-1块的结束字母相同// f[i][j] = min(f[i][j],f[i-1][k] + counts[i] - 1);// // 否则// // f[i][j] = min(f[i][j],f[i-1][k] + counts[i]);//// 第一种情况的开始和结束字母相同有个前提：// f[i-1][k]中的k在第i块中出现//// 之后的条件是：// 1)第i块只有一种字符// 2)第i块结束字符与第i-1块结束字符不相同//// ps：第一种情况是第二种情况的特殊，因为每种字母要么是在开始，// 要么是在结束，不会连续的两个块结束字母相同，这样肯定不是// 最优的，而如果第i块只有一种字符，那么显然不在这之列。//// wrong answer了10次，如此顽强我也是醉了//// 继续练吧#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);const int maxn = 1008;const int inf = 0x3f3f3f3f;char s[maxn];bool vis[maxn][30];int f[maxn][300];int n,k;int counts[maxn];void dp(){for (int i=0;i<=26;i++){if (vis[1][i]){f[1][i] = counts[1];}}for (int i=2;i<=n/k;i++){for (int j=0;j<26;j++){if (vis[i][j]){for (int m=0;m<26;m++){if (vis[i][m] && (counts[i]==1 || j!=m)){f[i][j] = min(f[i][j],f[i-1][m] + counts[i] - 1);}else{f[i][j] = min(f[i][j],f[i-1][m] + counts[i]);}}}}}int ans = inf;for (int i=0;i<26;i++){ans = min(ans,f[n/k][i]);}printf("%d\n",ans);}void print(){for (int i=1;i<=n/k;i++){cout << counts[i] << " ";}cout << endl;}void init(){scanf("%d %s",&k,s+1);n = strlen(s+1);memset(vis,0,sizeof(vis));memset(f,inf,sizeof(f));memset(counts,0,sizeof(counts));int x=1;for (int i=1;i<=n;i++){if (!vis[x][s[i]-'a']){counts[x]++;vis[x][s[i]-'a'] = 1;}if (i%k==0){x++;}}//cout << "x = " << x << endl;//print();dp();}int main() {int t;//freopen("E:\\Code\\1.txt","r",stdin);scanf("%d",&t);while(t--){init();}return 0;}