CodeForces - 327AFlipping Game

最新推荐文章于 2021-08-10 10:20:25 发布

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#codeforces

介绍了一个翻转游戏问题，目标是在限定时间内通过翻转区间内的0和1来获得最多数量的1。文章提供了两种解决方案，一种是使用暴力枚举区间的方法进行求解，另一种则是通过观察并利用前缀和来优化算法，最终实现O(n)的时间复杂度。

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A. Flipping Game

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Iahub got bored, so he invented a game to be played on paper.

He writes n integers a1, a2, …, an. Each of those integers can be either 0 or 1. He’s allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.

The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, …, an. It is guaranteed that each of those n values is either 0 or 1.

Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.

Examples
input
5
1 0 0 1 0
output
4
input
4
1 0 0 1
output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.

题意：

可以对一段区间里的01取反一次，n<=100，
1.暴力枚举区间在计算答案O(n^3)

2.暴力枚举区间前缀和优化算答案O(n^2)

3.观察每次用前缀和计算[L,R]答案(ai记录1-i中，1的个数,bi记录1-i中，0的个数）
ans=a[L-1]+a[n]-a[R]+a[R]-b[L-1];
现在考虑以R为右端点的区间，a[n]-a[R]+b[R]这一部分始终是相同的，所以我们只要找到a[L-1]-b[l-1]这段区间的最大值即可，而这个最大值可以直接开一个变量维护
复杂度O(n)

O(n^2)代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=1e5+5;
int a[maxn],b[maxn],n,x;

int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=1;i<=n;++i)
    {
      scanf("%d",&x);
      a[i]=a[i-1]+(x==1);
      b[i]=b[i-1]+(x==0);
    }
    int ans=0;
    for(int i=1;i<=n;++i)
      for(int j=i;j<=n;++j)
      ans=max(ans,a[i-1]+a[n]-a[j]+b[j]-b[i-1]);

    //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]);

    printf("%d\n",ans);
  }
  return 0;
}

O(n)代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=1e5+5;
int a[maxn],b[maxn],n,x;

int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=1;i<=n;++i)
    {
      scanf("%d",&x);
      a[i]=a[i-1]+(x==1);
      b[i]=b[i-1]+(x==0);
    }
    int ans=0,res=0;
    for(int i=1;i<=n;++i)
    {
      ans=max(ans,res+a[n]-a[i]+b[i]);
      if(a[i]-b[i]>res)res=a[i]-b[i];
    }


    //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]);

    printf("%d\n",ans);
  }
  return 0;
}