CF 220C Little Elephant and Shifts

题目链接:http://codeforces.com/problemset/problem/220/C

题目大意:

a为长度为n的不变的串,b为长度为n的可循环左移的串.

a,b均由1~n组成,每个数在每串中用一次.

a串与b串的值为min{ |i-j| },ai==bj

求所有bi,bi+1...bn,bn-1...b1,与a串的值.

题目思路:

1<=n<=100000,朴素算法O(n^2)肯定不行,所以想到了O(nlogn)的线段树,纠结了很久,搞出个维护绝对值的最小值,但是表示没有任何想法= =...

之后发现对于每次操作,要么是某些数+1,要么是某些数-1(首个数除外)...本来还想用线段树来写,写着写着发现怎么看怎么像优先队列+延迟标记...

代码:

#include <stdlib.h>#include <string.h>#include <stdio.h>#include <ctype.h>#include <math.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <string>#include <iostream>#include <algorithm>using namespace std;#define ll long long#define ls rt<<1#define rs ls|1#define lson l,mid,ls#define rson mid+1,r,rs#define middle (l+r)>>1#define eps (1e-9)#define type int#define clr_all(x,c) memset(x,c,sizeof(x))#define clr(x,c,n) memset(x,c,sizeof(x[0])*(n+1))#define MOD 1000000007#define inf 0x3f3f3f3f#define pi acos(-1.0)template <class T> void _swap(T &x,T &y){T t=x;x=y;y=t;}template <class T> T _max(T x,T y){return x>y? x:y;}template <class T> T _min(T x,T y){return x<y? x:y;}int test,cas;const int M=100000 +5;int n,m;int ap[M],b[M],cur[M];struct node{int id,cur,num;node(){}node(int _i,int _c,int _n){id=_i,cur=_c,num=_n;}bool operator < (const node& t) const{return num > t.num;}};priority_queue<node>rgt,lft;int covr,covl;void run(){int i,j;for(i=1;i<=n;i++){scanf("%d",&j);ap[j]=i;}while(!rgt.empty()) rgt.pop();while(!lft.empty()) lft.pop();clr(cur,0,n+1);for(i=1;i<=n;i++){scanf("%d",&b[i]);if(i>ap[b[i]]) rgt.push(node(b[i],0,i-ap[b[i]]));else lft.push(node(b[i],0,ap[b[i]]-i));}covr=covl=0;node r,l;for(i=1;;i++){while(!rgt.empty()){r=rgt.top();if(r.cur!=cur[r.id]) rgt.pop();else if(r.num+covr==0){rgt.pop();lft.push(node(r.id,r.cur,-covl));}else break;}while(!lft.empty()){l=lft.top();if(l.cur!=cur[l.id]) lft.pop();else break;}int ans=inf;if(!rgt.empty()) ans=_min(ans,rgt.top().num+covr);if(!lft.empty()) ans=_min(ans,lft.top().num+covl);printf("%d\n",ans);if(i==n) break;covr--,covl++;cur[b[i]]++;if(n>ap[b[i]]) rgt.push(node(b[i],cur[b[i]],n-ap[b[i]]-covr));else lft.push(node(b[i],cur[b[i]],-covl));}}void preSof(){}int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);preSof();//run();while(~scanf("%d",&n)) run();//for(scanf("%d",&test),cas=1;cas<=test;cas++) run();return 0;}