D - Handstand
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400400 points
Problem Statement
NN people are arranged in a row from left to right.
You are given a string SS of length NN consisting of 0 and 1, and a positive integer KK.
The ii-th person from the left is standing on feet if the ii-th character of SS is 0, and standing on hands if that character is 1.
You will give the following direction at most KK times (possibly zero):
Direction: Choose integers ll and rr satisfying 1≤l≤r≤N1≤l≤r≤N, and flip the ll-th, (l+1)(l+1)-th, ......, and rr-th persons. That is, for each i=l,l+1,...,ri=l,l+1,...,r, the ii-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.
Find the maximum possible number of consecutive people standing on hands after at most KK directions.
Constraints
- NN is an integer satisfying 1≤N≤1051≤N≤105.
- KK is an integer satisfying 1≤K≤1051≤K≤105.
- The length of the string SS is NN.
- Each character of the string SS is
0or1.
Input
Input is given from Standard Input in the following format:
NN KK SS
Output
Print the maximum possible number of consecutive people standing on hands after at most KK directions.
Sample Input 1 Copy
Copy
5 1 00010
Sample Output 1 Copy
Copy
4
We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:
- Give the direction with l=1,r=3l=1,r=3, which flips the first, second and third persons from the left.
Sample Input 2 Copy
Copy
14 2 11101010110011
Sample Output 2 Copy
Copy
8
Sample Input 3 Copy
Copy
1 1 1
Sample Output 3 Copy
Copy
1
No directions are necessary.
题意：
给你一个只含有0和1的字符串，并且给你一个数字K，你可以选择最多K次区间，每一个区间L<=R，然后对这个区间的元素进行取反操作。
即0变成1，1变成0，问你在最聪明的操作之后最大可以获得的连续1的串是多长？
可以看样例解释理解题意。
思路：
我们对字符串的连续1和0串进行计数处理，即把连续的1或者0的个数记录起来，
那么字符串会生成如下的数组
例如字符串是 110000111001101
我们把连续的1和连续0分别放入a和b数组，
那么a的元素是2 3 2 1
b的元素是4 2 1
我们思考可以发现，我们想要最长的连续1串，那么我们处理的区间肯定是连续的0区间，让他们变成1，然后来增长连续1串的长度。
即处理的0区间一定是连续的。
那么我们可以知道如下，例如我们处理两个区间，那么可以的得到的最长的连续1串就是这两个区间的0串长度的sum和以及这两个0串前后和中间的1串的sum和。
那么我们只需要枚举连续的K个的0串即可，
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ‘\0‘, sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=100010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int k;
char s[maxn];
std::vector<int> v1,v2;
ll sum1[maxn];
ll sum2[maxn];
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
gbtb;
cin>>n>>k;
cin>>s;
int cnt=0;
// int i=0;
if(s[0]==‘0‘)
{
v1.pb(0);
}
repd(i,0,n-1)
{
if(s[i]==‘0‘)
{
while(s[i]==‘0‘)
{
cnt++;
i++;
}
v2.push_back(cnt);
cnt=0;
i--;
}
if(s[i]==‘1‘)
{
while(s[i]==‘1‘)
{
cnt++;
i++;
}
v1.push_back(cnt);
cnt=0;
i--;
}
}
int num=max(sz(v1),sz(v2));
repd(i,1,maxn-2)
{
v1.push_back(0);
v2.push_back(0);
}
repd(i,1,maxn-2)
{
sum1[i]+=sum1[i-1]+v1[i-1];
sum2[i]+=sum2[i-1]+v2[i-1];
}
int w=0;
ll ans=0ll;
repd(i,1,n-k+4)
{
int l=i;
int r=l+k;
ll temp=sum1[r]-sum1[l-1];
temp+=sum2[r-1]-sum2[l-1];
ans=max(ans,temp);
}
cout<<ans<<endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ‘ ‘ || ch == ‘\n‘);
if (ch == ‘-‘) {
*p = -(getchar() - ‘0‘);
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 - ch + ‘0‘;
}
}
else {
*p = ch - ‘0‘;
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 + ch - ‘0‘;
}
}
}
原文地址：https://www.cnblogs.com/qieqiemin/p/10704797.html