A boy named Leo doesn't miss a single TorCoder contest round. On the last TorCoder round number 100666 Leo stumbled over the following problem. He was given a string s, consisting of n lowercase English letters, and m queries. Each query is characterised by a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
We'll consider the letters in the string numbered from 1 to n from left to right, that is, s = s1s2... sn.
After each query he must swap letters with indexes from li to ri inclusive in string s so as to make substring (li, ri) a palindrome. If there are multiple such letter permutations, you should choose the one where string (li, ri) will be lexicographically minimum. If no such permutation exists, you should ignore the query (that is, not change string s).
Everybody knows that on TorCoder rounds input line and array size limits never exceed 60, so Leo solved this problem easily. Your task is to solve the problem on a little bit larger limits. Given string s and m queries, print the string that results after applying all m queries to string s.
The first input line contains two integers n and m (1 ≤ n, m ≤ 105) — the string length and the number of the queries.
The second line contains string s, consisting of n lowercase Latin letters.
Each of the next m lines contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n) — a query to apply to the string.
In a single line print the result of applying m queries to string s. Print the queries in the order in which they are given in the input.
7 2
aabcbaa
1 3
5 7
abacaba
3 2
abc
1 2
2 3
abc
A substring (li, ri) 1 ≤ li ≤ ri ≤ n) of string s = s1s2... sn of length n is a sequence of characters slisli + 1...sri.
A string is a palindrome, if it reads the same from left to right and from right to left.
String x1x2... xp is lexicographically smaller than string y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or exists such number r(r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1.
对于每个字母开一个线段树维护区间
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code:
// #include<bits/stdc++.h> using namespace std; #define maxnn 600000 struct node { int tot,a,b,laz; }tree[maxnn][27]; int cnt[40]; int n,m; void pushdown(int p,int k) { tree[p<<1][k].laz=tree[p][k].laz; tree[p<<1][k].tot=(tree[p<<1][k].b-tree[p<<1][k].a+1)*tree[p][k].laz; tree[p<<1|1][k].laz=tree[p][k].laz; tree[p<<1|1][k].tot=(tree[p<<1|1][k].b-tree[p<<1|1][k].a+1)*tree[p][k].laz; tree[p][k].laz=-1; } void pushdown1(int p,int k) { tree[p][k].laz=-1; tree[p<<1][k].laz=0; tree[p<<1][k].tot=0; tree[p<<1|1][k].laz=0; tree[p<<1|1][k].tot=0; tree[p][k].laz=-1; } void query(int p,int x,int y,int k,int d) { if(tree[p][k].laz==0) pushdown1(p,k); if(tree[p][k].laz!=0&&tree[p][k].laz!=-1) pushdown(p,k); if(tree[p][k].a>=x&&tree[p][k].b<=y) { tree[p][k].laz=d; tree[p][k].tot=(tree[p][k].b-tree[p][k].a+1)*d; } else { int mid=(tree[p][k].a+tree[p][k].b)>>1; if(x<=mid&&y>=tree[p][k].a) { query(p<<1,x,y,k,d); } { if(x<=tree[p][k].b&&y>mid) query(p<<1|1,x,y,k,d); } tree[p][k].tot=tree[p<<1][k].tot+tree[p<<1|1][k].tot; } } int sum(int p,int x,int y,int k) { if(tree[p][k].laz==0) pushdown1(p,k); if(tree[p][k].laz!=0&&tree[p][k].laz!=-1) pushdown(p,k); if(tree[p][k].a>=x&&tree[p][k].b<=y) { return tree[p][k].tot; } else { int mid=(tree[p][k].a+tree[p][k].b)>>1; int l=0,r=0; if(x<=mid&&y>=tree[p][k].a) { l=sum(p<<1,x,y,k); } { if(x<=tree[p][k].b&&y>mid) r=sum(p<<1|1,x,y,k); } int tot=l+r; return tot; } } int ccc[40],tot; void qing(int p,int x,int y,int k,int d) { if(tree[p][k].laz!=0&&tree[p][k].laz!=-1) pushdown(p,k); if(tree[p][k].laz==0) pushdown1(p,k); if(tree[p][k].a>=x&&tree[p][k].b<=y) { tree[p][k].tot=0; tree[p][k].laz=0; } else { int mid=(tree[p][k].a+tree[p][k].b)/2; if(x<=mid&&y>=tree[p][k].a) { qing(p<<1,x,y,k,d); } if(x<=tree[p][k].b&&mid<y) { qing(p<<1|1,x,y,k,d); } tree[p][k].tot=tree[p<<1][k].tot+tree[p<<1|1][k].tot; } } void build(int x,int y,int p,int k) { tree[p][k].a=x; tree[p][k].b=y; tree[p][k].laz=-1; if(x<y) { build(x,(x+y)>>1,p<<1,k); build((x+y)/2+1,y,p<<1|1,k); } else tree[p][k].tot=0; return ; } int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); cin>>n>>m; char T; int x,y; for(int i=1;i<=26;i++) { build(1,n,1,i); } scanf("%c",&T); for(int i=1;i<=n;i++) { scanf("%c",&T); query(1,i,i,int(T-'a'+1),1); } while(m--) { memset(cnt,0,sizeof(cnt)); int tott=0; int ji=0; int jichu=0; cin>>x>>y;tot=0; for(int i=1;i<=26;i++) { cnt[i]=sum(1,x,y,i); if(cnt[i]&1) { ji++; jichu=i; } tott+=cnt[i]; } if(ji>1) { continue; } else { for(int j=1;j<=26;j++) qing(1,x,y,j,0); if(ji==1) { int l=x,r=y; cnt[jichu]--; query(1,x+(tott)/2,x+(tott)/2,jichu,1); for(int i=1;i<=26;i++) { if(cnt[i]) { query(1,l,l+cnt[i]/2-1,i,1); query(1,r-cnt[i]/2+1,r,i,1); l=l+cnt[i]/2; r=r-cnt[i]/2; } } } if(ji==0) { int tmp=0; int l=x,r=y; for(int i=1;i<=26;i++) { if(cnt[i]) { query(1,l,l+cnt[i]/2-1,i,1); query(1,r-cnt[i]/2+1,r,i,1); l=l+cnt[i]/2; r=r-cnt[i]/2; } } } } } for(int i=1;i<=n;i++) { for(int j=1;j<=26;j++) { if(sum(1,i,i,j)) { printf("%c",(char)(j+'a'-1)); } } } }