CodeForces-1680C Binary String
Binary String
二分 + 尺取 || 尺取
二分+尺取:时间复杂度为 \(O(nlogn)\)
答案是单调的,所以直接二分枚举答案,然后再 judge 判断的时候,尺取中间剩下的区间
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <functional>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <deque>
#include <stack>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
string s;
int sum = 0;
int query(int x)
{
int l = 0, r = 0, len = s.length(), cnt0 = 0, cnt1 = 0;
while(l < len)
{
while(r < len && cnt0 <= x)
{
if(s[r] == '0') cnt0++;
else cnt1++;
r++;
}
if(sum - cnt1 <= x) return true;
if(s[l] == '1') cnt1--;
else cnt0--;
l++;
}
return max(cnt0, sum - cnt1) <= x;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while(t--)
{
cin >> s;
int l = 0, r = s.length();
sum = 0;
for(int i=0; i<s.length(); i++) if(s[i] == '1') sum++;
while(l < r)
{
int mid = l + r >> 1;
if(query(mid))
r = mid;
else
l = mid + 1;
}
cout << l << endl;
}
return 0;
}
尺取,时间复杂度为 \(O(n)\)
这里用到一个贪心,只有在剩余区间的代价和删除区间的代价尽可能相等的时候,是最优解
所以就可以根据这个,尺取每个区间 \([l, r]\),因为对于区间 \([l_{i+1}, r_{i+1}]\),必然有 \(r_i \leq r_{i+1}\)
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while(t--)
{
string s;
cin >> s;
int cnt0 = 0, cnt1 = 0, len = s.length();
for(int i=0; i<len; i++) cnt1 += s[i] == '1';
int l = 0, r = 0;
int ans = len;
while(l < len)
{
while(r < len && cnt1 != cnt0)
{
if(s[r] == '0') cnt0++;
else cnt1--;
r++;
}
ans = min(ans, max(cnt0, cnt1));
if(s[l] == '0') cnt0--;
else cnt1++;
l++;
}
cout << ans << endl;
}
return 0;
}