codeforces Flipping Game 题解

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.

Sample test(s)

input

5

1 0 0 1 0

output

本题由于数据量小，能够使用暴力法，时间效率是O(n^3)

可是这里巧用最大子段和的思想。能够把时间效率降到O(n)

思想：

1 想使用一个新的数列，计算连续出现了多少个1和连续出现了多少个零

2 求这个新数列的最大子段和

3 Flip最大子段中的 0 和 1,

4 计算出结果

比暴力法复杂非常多了，可是时间效率却提高了三个档次。

#include <vector>

#include <string>

#include <iostream>

using namespace std;

void FlippingGame()

{

	int n, a;

	cin>>n;

	vector<bool> vbn(n);

	for (int i = 0; i < n; i++)

	{

		cin>>a;

		vbn[i] = a;

	}

	vector<int> ans;

	int c = 1;

	for (int i = 1; i < n; i++)

	{

		if (vbn[i] == vbn[i-1]) c++;

		else

		{

			if (vbn[i-1]) ans.push_back(-c);

			else ans.push_back(c);

			c = 1;

		}

	}

	if (vbn.back()) ans.push_back(-c);

	else ans.push_back(c);

	//求最大子段和思想

	int stTmp = 0, st = ans.size(), end = ans.size(), maxVal = 0, sum = 0;

	for (unsigned i = 0; i < ans.size(); i++)

	{

		sum += ans[i];

		if (sum > maxVal)

		{

			st = stTmp;

			maxVal = sum;

			end = i;

		}

		if (sum <= 0)

		{

			sum = 0;

			stTmp = i+1;

		}

	}

	int nums = 0;

	for (int i = 0; i < ans.size(); i++)

	{

		if (ans[i] < 0) nums += ans[i];

	}

	if (maxVal > 0) cout<<maxVal - nums;

	else cout<<-(ans.front()+1);

}

巴特西

codeforces Flipping Game 题解

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