codeforces 558 E A Simple Task
做法的話就是用線段樹維護區間和
然後,在考慮字符‘a‘和‘b‘的情況,操作的情況和上面類似,字符‘a‘和‘b‘出現的位置值為1。否則為0,q次操作後,假設一個位置的值為1。而且該位置沒有填寫‘a‘。那麽這個位置就填寫‘b‘。
接下來的情況以此類推,考慮abc。abcd,......,abcd...z。
時間復雜度O(26*(n+q)logn)。
#include<map> #include<string> #include<cstring> #include<cstdio> #include<cstdlib> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<climits> #include<list> #include<iomanip> #include<stack> #include<set> using namespace std; struct Tree { int l,r,sum; }tree[int(4e5)]; void create(int l,int r,int k) { tree[k].l=l; tree[k].r=r; tree[k].sum=0; if(l==r) return; int m=l+r>>1; create(l,m,k<<1); create(m+1,r,k<<1|1); } void lazy(int k) { if(tree[k].l!=tree[k].r&&tree[k].sum!=tree[k<<1].sum+tree[k<<1|1].sum) { if(tree[k].sum==0) tree[k<<1].sum=tree[k<<1|1].sum=0; else { tree[k<<1].sum=tree[k<<1].r-tree[k<<1].l+1; tree[k<<1|1].sum=tree[k<<1|1].r-tree[k<<1|1].l+1; } } } void update(int l,int r,bool flag,int k) { lazy(k); if(l==tree[k].l&&r==tree[k].r) { tree[k].sum=flag?r-l+1:0; return; } int m=tree[k].l+tree[k].r>>1; if(r<=m) update(l,r,flag,k<<1); else if(l>m) update(l,r,flag,k<<1|1); else { update(l,m,flag,k<<1); update(m+1,r,flag,k<<1|1); } tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum; } int seek(int l,int r,int k) { lazy(k); if(l==tree[k].l&&r==tree[k].r) return tree[k].sum; int m=tree[k].l+tree[k].r>>1; if(r<=m) return seek(l,r,k<<1); if(l>m) return seek(l,r,k<<1|1); return seek(l,m,k<<1)+seek(m+1,r,k<<1|1); } void update(int l,int r,bool flag) { int t=seek(l,r,1); if(flag) { if(l<=l+t-1) update(l,l+t-1,1,1); if(l+t<=r) update(l+t,r,0,1); } else { if(r-t+1<=r) update(r-t+1,r,1,1); if(l<=r-t) update(l,r-t,0,1); } } struct Box { int l,r; bool flag; }box[int(1e5)+10]; int ans[int(1e5)+10]; int main() { int n,q; cin>>n>>q; create(1,n,1); string s; cin>>s; for(int i=0;i<q;i++) cin>>box[i].l>>box[i].r>>box[i].flag; for(int i=0;i<26;i++) { update(1,n,0,1); for(int j=0;j<n;j++) if(s[j]<=i+‘a‘) update(j+1,j+1,1,1); for(int j=0;j<q;j++) update(box[j].l,box[j].r,box[j].flag); for(int j=0;j<n;j++) if(ans[j]==0&&seek(j+1,j+1,1)==1) ans[j]=i+‘a‘; } for(int i=0;i<n;i++) putchar(ans[i]); }
time limit per test 5 seconds memory limit per test 512 megabytes input standard input output standard output
This task is very simple. Given a string S of length n and q queries
each query is on the format i j k which
means sort the substring consisting of the characters from i
Output the final string after applying the queries.
InputThe first line will contain two integers n,?q (1?≤?n?≤?105, 0?≤?q?≤?50?000), the length of the string and the number of queries respectively.
Next line contains a string S itself. It contains only lowercase English letters.
Next q lines will contain three integers each i,?j,?k (1?≤?i?≤?j?≤?n, ).
OutputOutput one line, the string S after applying the queries.
Sample test(s) input10 5 abacdabcda 7 10 0 5 8 1 1 4 0 3 6 0 7 10 1output
cbcaaaabddinput
10 1 agjucbvdfk 1 10 1output
abcdfgjkuvNote
First sample test explanation:
codeforces 558 E A Simple Task