标签：style blog class code ext color
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);
}
codeforces Flipping Game 题解,布布扣,bubuko.com
标签：style blog class code ext color
原文地址：http://blog.csdn.net/kenden23/article/details/24862445
