POJ 3185 模拟加开关

从左到右,从右到左,以此来一遍尺取法,因为一开始如果第0位为1的话,我们是取得第一位反转,使得0,1,2三位被反转,就略去了只是反转0,1的情况了

#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 a[21];
int vis[21];
int main()
{
 up(i, 0, 20)
 {
  cin >> a[i];
 }
 memcpy(vis, a, sizeof(a));
 int cnt2= 0;
 int cnt1 = 0;
 up(i, 0, 20)
 {
  if (a[i] == 1)
  {
   cnt1++;
   if (i == 19)
   {
    cnt1 = 200;
    break;
   }
   a[i + 1] = !a[i + 1];
   a[i + 2] = !a[i + 2];
  }
 }
 dwd(i, 19, 0)
 {
  if (vis[i]==1)
  {
   cnt2++;
   if (i == 0)
   {
    cnt1 = 200;
    break;
   }
   vis[i - 1] = !vis[i - 1];
   vis[i - 2] = !vis[i - 2];
  }
 }
 cout << min(cnt1, cnt2);
 return 0;
}

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转载自blog.csdn.net/weixin_44019404/article/details/88649576