HNOI2018 转盘

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结论:一定存在某种最优方案满足先在起点停留一段时间,然后开始走一圈。否则一定能转化。

把序列复制一份在后面。枚举起点$s$,那么答案就是$\min\limits_{s\le n}(\max\limits_{i=0}^{n-1}(a_{i+s}-i)+n-1)$

设$w_i=a_i-i$,则答案就变成了$\min\limits_{s\le n}(s+\max\limits_{i=0}^{n-1}w_{i+s})+n-1$。

因为$w_{i+n}=w_i-n$,所以取max把$w_{s+n}$及之后的加上没有影响。

答案就是$\min\limits_{s\le n}(s+\max\limits_{s\le i}w_{i})+n-1$

然后就是类似于楼房重建这样维护。

时间复杂度$O((n+m)\log ^2n)$

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/*
Author: CNYALI_LK
LANG: C++
PROG: 4425.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline void read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}

inline void read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (char &x) {
x=gc();
}
inline void read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r');
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
}
template<typename A,typename ...B>
inline void read(A &x,B &...y){
read(x);read(y...);
}
// print a signed integer
inline void write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}

inline void write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}
inline void write (char x) {
putc(x);
}
inline void write(const char *x){
while(*x){putc(*x);++x;}
}
inline void write(char *x){
while(*x){putc(*x);++x;}
}
template<typename A,typename ...B>
inline void write(A x,B ...y){
write(x);write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int n,m,p,a[200005],lsa;
struct smt{
int ans,ansall,mx;
int ls,rs;
smt *l,*r;
int calc(int xmx){
if(ls==rs)return ls+max(mx,xmx);
if(xmx>r->mx)return min(l->calc(xmx),r->ls+xmx);
else return min(r->calc(xmx),ans);
}
void push_up(){
mx=max(l->mx,r->mx);
ans=l->calc(r->mx);
ansall=min(ans,r->ansall);
}
smt(int la,int ra){
ls=la;rs=ra;
if(ls==rs){
mx=a[ls];
ans=ansall=a[ls]+ls;
}
else{
int mid=(ls+rs)>>1;
l=new smt(ls,mid);
r=new smt(mid+1,rs);
push_up();
}
}
void update(int x,int y){
if(ls==rs){mx=y;ans=ansall=x+y;}
else{
if(x<=l->rs)l->update(x,y);
else r->update(x,y);
push_up();
}
// write("...",ls,' ',rs,':',ans,',',ansall,'\n');
}
};
smt *r;
int main(){
#ifdef cnyali_lk
freopen("4425.in","r",stdin);
freopen("4425.out","w",stdout);
#endif
read(n,m,p);
for(int i=1;i<=n;++i){read(a[i]);a[i]-=i;a[i+n]=a[i]-n;}
r=new smt(1,2*n);
write(lsa=r->ans+n-1,'\n');
int x,y;
for(;m;--m){
read(x,y);
if(p){x^=lsa;y^=lsa;}
a[x]=y-x;a[x+n]=a[x]-n;
r->update(x,a[x]);
r->update(x+n,a[x+n]);
// for(int i=1;i<=n+n;++i)write(a[i],i==n+n?'\n':' ');
// write(r->ans,' ',r->ansall,'\n');
write(lsa=r->ans+n-1,'\n');
}
return 0;
}
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