CF724D. Dense Subsequence[贪心 字典序!]
You are given a string s, consisting of lowercase English letters, and the integer m.
One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.
Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.
Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.
Find the lexicographically smallest string, that can be obtained using this procedure.
The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).
The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.
Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.
3
cbabc
a
2
abcab
aab
3
bcabcbaccba
aaabb
In the first sample, one can choose the subsequence {3} and form a string "a".
In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".
题意:至少每m个选一个字符使得选择的字符字典序最小的排列的字典序最小
和那个经典问题很想,就是多了选择的字符可以任意排列
要深入理解字典序
考虑只选a,可行就只选a,否则就要选b,此时a一定全选(字典序最小),b尽量少的选
枚举选到哪个字符就行了
PS:貌似可以二分,然而才26个字符并不需要
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N=1e5+5,INF=1e9+5; int n,m,a[300]; char s[N]; bool sol(char c){ int last=0,cnt=0,lc=0; for(int i=1;i<=n;i++){ if(s[i]==c) lc=i; if(s[i]<c) last=i; if(i-last>=m){ if(i-lc<m) last=lc,cnt++; else return false; } } for(int i=1;i<=n;i++) a[s[i]]++; for(char now='a';now<c;now++){ while(a[now]--) putchar(now); } while(cnt--) putchar(c); return true; } int main(){ scanf("%d%s",&m,s+1); n=strlen(s+1); for(char c='a';c<='z';c++) if(sol(c)) break; }