2 seconds
256 megabytes
standard input
standard output
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as anot-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and{1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
8
10000011
5
2
01
2
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
题意：给出1个01串，必须将某一段取反，问选择一个1、0间隔的子序列，最长有多长？
分析：
我真是傻逼。
我想了个dp，dp[i]表示前i个中，取反的那个串最后一个数是第i个，的答案。
预处理两个数组，front[i]表示不修改的前i个的答案，back[i]表示不修改的i~n的答案。
那么转移显然
dp[i] = max(dp[i - 1] + (arr[i] ^ arr[i - 1]), front[i - 1] + (arr[i] == arr[i - 1]));
前一个表示修改的是一段的情况，后一个代表只修改第i个的情况。
答案为
ans = max(ans, dp[i] + back[i + 1]);
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
int n;
string data;
bool arr[N];
int front[N], back[N], dp[N], ans; inline void Input()
{
scanf("%d", &n);
cin >> data;
} inline void Solve()
{
for(int i = ; i < n; i++) arr[i + ] = data[i] == '';
int cnt[] = {};
arr[n + ] = arr[n] ^ ;
for(int i = n; i > ; i--)
{
if(arr[i + ] ^ arr[i]) cnt[arr[i]] = cnt[arr[i] ^ ] + ;
back[i] = cnt[arr[i - ]];
}
cnt[] = cnt[] = , arr[] = arr[] ^ ;
for(int i = ; i < n; i++)
{
if(arr[i] != arr[i - ]) cnt[arr[i]] = cnt[arr[i] ^ ] + ;
front[i] = cnt[arr[i + ]];
} for(int i = ; i <= n; i++)
{
dp[i] = max(dp[i - ] + (arr[i] ^ arr[i - ]),
front[i - ] + (arr[i] == arr[i - ]));
ans = max(ans, dp[i] + back[i + ]);
}
printf("%d\n", ans);
} int main()
{
freopen("a.in", "r", stdin);
Input();
Solve();
return ;
}
可是正解简单到令人发指。
#include <bits/stdc++.h>
using namespace std; int N, res = ;
string S; int main(){
cin >> N >> S;
for(int i = ; i < N; i++){
res += (S[i] != S[i - ]);
}
cout << min(res + , N) << '\n';
}
其实很容易理解，修改比不修改的答案最多增加2，举几个例子就知道了。。。