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Language: Default Face The Right Way
Description Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect. Fortunately, FJ recently bought an automatic cow turning machine. Since he purchased the discount model, it must be irrevocably preset to turn K (1 ≤ K ≤ N) cows at once, and it can only turn cows that are all standing Because FJ must pick a single, never-changing value of K, please help him determine the minimum value of K that minimizes the number of operations required by the machine to make all the cows face forward. Also determine M, the Input Line 1: A single integer: N Lines 2..N+1: Line i+1 contains a single character, F or B, indicating whether cow i is facing forward or backward. Output Line 1: Two space-separated integers: K and M Sample Input 7 B B F B F B B Sample Output 3 3 Hint For K = 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7) Source |
题意：n头牛站成一列，每头牛或者向前或者向后，要求通过转向让所有的牛都面向前方，使K头连续的牛转向为一次操作，求最小的操作次数M和对应的K。开关问题
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 5005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;
int dir[maxn]; //牛的方向
int f[maxn]; //区间[i,i+k-1]是否进行翻转
int n;
char str[5];
int calc(int k)
{
memset(f,0,sizeof(f));
int ans=0;
int sum=0; //f的和
for (int i=0;i+k<=n;i++)
{
//计算区间[i,i+k-1]
if ((sum+dir[i])%2!=0) //前端的牛朝后方
{
f[i]=1;
ans++;
}
sum+=f[i];
if (i-k+1>=0)
sum-=f[i-k+1];
}
//检查后方的牛是否有面朝后方的情况
for (int i=n-k+1;i<n;i++)
{
if ((sum+dir[i])%2!=0)
return -1;
if (i-k+1>=0)
sum-=f[i-k+1];
}
return ans;
}
void solve()
{
int K=1,M=n;
for (int i=1;i<=n;i++)
{
int m=calc(i);
if (m>=0&&m<M)
{
K=i;
M=m;
}
}
printf("%d %d\n",K,M);
}
int main()
{
while (~scanf("%d",&n))
{
memset(dir,0,sizeof(dir));
for (int i=0;i<n;i++)
{
scanf("%s",str);
if (str[0]=='B') dir[i]=1;
}
solve();
}
return 0;
}