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

51 0 0 1 0

output

4

本题因为数据量小，可以使用暴力法，时间效率是O(n^3)

但是这里巧用最大子段和的思想，可以把时间效率降到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

比暴力法复杂很多了，但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);}