尺取法和开关反转问题
这里最可以用暴力去递归发现时间复杂度大概为n^3左右。这显然会超时。
我们就用尺取法来取就好了。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const long long INF = 0x3f3f3f3f;
using namespace std;
typedef pair<int, int> pir;
int n;
int dir[5005];
int f[5005];
int solve(int k)
{
memset(f, 0, sizeof(f));
int res = 0;
int sum = 0;
upd(i, 0, n -k)
{
if ((sum+dir[i]) & 1)
{
res++;
f[i] = 1;
}
sum += f[i];
if (i - k+1 >= 0)
{
sum -= f[i - k+1];
}
}
up(i, n - k + 1, n)
{
if ((dir[i] + sum) & 1)return -1;
if(i-k+1>=0)
sum -= f[i - k + 1];
}
return res;
}
int main()
{
cin >> n;
char c;
up(i, 0, n)
{
cin >> c;
if (c == 'B')
dir[i] = 1;
else dir[i] = 0;
}
int ans=n, t=1;
upd(i, 1, n)
{
if (solve(i) == -1)continue;
else
{
if (ans > solve(i))
{
ans = solve(i);
t = i;
}
}
}
cout << t<<" "<< ans << endl;
return 0;
}