Dense Subsequence
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".
分析:贪心,在选取当前字符后最多隔m个取一个最小的(坐标尽量靠后);
这样选完后扫一遍字符串,那些没有被选取且不是当前答案里最大的字符也能加入答案;
最后排序即可;
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, ls[rt] #define Rson mid+1, R, rs[rt] const int maxn=2e5+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,k,t,vis[maxn]; struct node { char x; int pos; node(char _x,int _pos):x(_x),pos(_pos){} bool operator<(const node&p)const { return x==p.x?pos>p.pos:x<p.x; } }; char a[maxn],ma; string ans; set<node>pq; int main() { int i,j; scanf("%d%s",&n,a); int len=strlen(a),pre=0; for(i=0;i<n;i++)pq.insert(node(a[i],i)); i=n; while(1) { node now=*pq.begin(); ans+=now.x; ma=max(ma,now.x); vis[now.pos]=1; for(j=pre;j<=now.pos;j++)pq.erase(node(a[j],j)); pre=now.pos+1; int cnt=0; for(j=i;j<len&&j-now.pos<=n;j++) { i=j; cnt=j-now.pos; pq.insert(node(a[j],j)); } i++; if(cnt<n)break; } for(i=0;a[i];i++)if(!vis[i]&&a[i]<ma)ans+=a[i]; sort(ans.begin(),ans.end()); cout<<ans<<endl; //system("Pause"); return 0; }