标签：too test ati syn 思路 lse flip pair inf
D - Handstand
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400400 points
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.
0 or 1.Input is given from Standard Input in the following format:
NN KK SS
Print the maximum possible number of consecutive people standing on hands after at most KK directions.
5 1 00010
4
We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:
14 2 11101010110011
8
1 1 1
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‘; } } }
AtCoder Beginner Contest 124 D - Handstand（思维+前缀和）
标签：too test ati syn 思路 lse flip pair inf
原文地址：https://www.cnblogs.com/qieqiemin/p/10704797.html
