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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.
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.
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
5 1 0 0 1 0
4
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); }