杭电多校第三场-H-Game

题目描述

Again Alice and Bob is playing a game with stones. There are N piles of stones labelled from 1 to N, the i th pile has ai stones.

First Alice will choose piles of stones with consecutive labels, whose leftmost is labelled with L and the rightmost one is R. After, Bob will choose another consecutive piles labelled from l to r (L≤l≤r≤R). Then they're going to play game within these piles.

Here's the rules of the game: Alice takes first and the two will take turn to make a move: choose one pile with nonegetive stones and take at least one stone and at most all away. One who cant make a move will lose.

Bob thinks this game is not so intersting because Alice always take first. So they add a new rule, which is that Bob can swap the number of two adjacent piles' stones whenever he want before a new round. That is to say, if the i th and i+1 pile have ai and ai+1 stones respectively, after this swapping there will be ai+1 and ai.

Before today's game with Bob, Alice wants to know, if both they play game optimally when she choose the piles from L to R, there are how many pairs (l, r) chosed by Bob that will make Alice *win*.

输入

Input contains several test cases.

For each case:

The fisrt line contains N, M. N is mentioned aboved ans M is the number of the sum of game rounds and the times of Bob's swapping.

The second line contains N integars a1,a2,...an, indicating the number of each piles' stones.

The next M lines will have an integar opt (1≤opt≤2), indicating the type of operation.

If opt equals 1, then L and R follow. Alice and Bob start a new round and Alice choose L and R as mentioned.

If opt equals 2, then POS follows. Bob will swap the piles labelled POS and POS+1.

0≤ai≤1,000,000

1≤N,M≤100,000,∑N,∑M<600,000

1≤L≤R≤N

1≤POS<N

输出

For each case:

For each opt which equals 1, you shall output one line with an integar indicating the number of pairs (l,r) that will make Alice win the round.

题意：
看起来是个博弈，但是分析下来发现题意就是让你找[L,R]区间内有多少个[l,r]满足al^a(l+)^...^ar！=，那不就是莫队吗
然后发现还要支持修改，每次选择一个位置pos，将a[pos]和a[pos+]交换
那不就不会了吗

三维莫队，即带修改的莫队
在普通的莫队上增加一维版本号，对询问排序时，先按块排，再按版本号排
更新的时候，先更新版本，再更新区间

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e6+;
int T,n,m;
int block;
ll tot;
struct orz{
    int l,r,t,id;
    bool operator < (const orz &x) const
    {
        if (l/block==x.l/block)
        {
            if (r/block==x.r/block) return t<x.t;
            return r<x.r;
        }
        return  l/block<x.l/block;
    }
}p[N];
int a[N],val[N],c[N],sum[N*];
ll ans[N];
inline int read()
{
    int x=;char ch=getchar();
    while(ch<''||ch>''){ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x;
}
void Go(int l,int r,int i)
{
    int pos=c[i]; //cout<<pos<<endl;
    if (pos>=l&&pos<=r)
    {
        sum[val[pos]]--;
        tot-=sum[val[pos]];
    }
    val[pos]^=a[pos]; val[pos]^=a[pos+];
    if (pos>=l&&pos<=r)
    {
        tot+=sum[val[pos]];
        sum[val[pos]]++;
    }
    swap(a[pos],a[pos+]);
    //for (int i=1;i<=n;i++) cout<<val[i]<<' '; cout<<endl;
}

void add(int x)
{
    tot+=sum[x];sum[x]++;
}
void del(int x)
{
    sum[x]--;tot-=sum[x];
}
int main()
{
 //   freopen("00.in","r",stdin);
  //  freopen("1.out","w",stdout);
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        int ti=,cnt=;
        for (int i=;i<=n;i++) a[i]=read(),val[i]=val[i-]^a[i];
        //for (int i=1;i<=n;i++) cout<<val[i]<<' '; cout<<endl;
        int op,x,y;
        for (int i=;i<=m;i++)
        {
            op=read();
            if (op==)
            {
                x=read(); y=read();
                p[++cnt].l=x-; p[cnt].r=y; p[cnt].t=ti; p[cnt].id=cnt;
            }
            else
            {
                c[++ti]=read();
            }
        }

        block=pow(n,2.0/3.0);
        sort(p+,p++cnt);

      //  sum[0]++;
        int l=,r=,t=; tot=;
        for(int i=;i<=cnt;i++)
        {
            while (t<p[i].t) Go(l,r,++t);
            while (t>p[i].t) Go(l,r,t--);

            //cout<<t<<endl;
            while (l>p[i].l) add(val[--l]);
            while (l<p[i].l) del(val[l++]);
            while (r<p[i].r) add(val[++r]);
            while (r>p[i].r) del(val[r--]);
            ll len=p[i].r-p[i].l; //cout<<len<<' '<<tot<<endl;
            ans[p[i].id]=len*(len-)/+len-tot;
        }

        for (int i=;i<=cnt;i++) printf("%lld\n",ans[i]);
        for (int i=;i<=n;i++) sum[val[i]]=;
    }
  //  fclose(stdin);
  //  fclose(stdout);
    return ;
}