c – 具有延迟传播时间限制问题的段树

#include <iostream>
#include <iostream>
#include <cstdio>
#include <cstring>
#define MAX 100000
using namespace std;
int M[2*MAX+1];
int flag[2*MAX+1];
int count;
void refresh(int begin,int end,int n)
{
    M[n] = end-begin+1 - M[n];
    flag[n]=0;
    flag[n*2] =!flag[n*2];
    flag[n*2+1] =!flag[n*2+1];
}
void update(int begin,int i,int j,int n=1)
{
    if(flag[n])
    {
        refresh(begin,end,n);
    }
    if(begin>=i && end<=j)
    {
        if(!flag[n])
        {
            refresh(begin,n);
        }
        flag[n] = 0;
        return;
    }
    else if(begin>=end)
    {
        return;
    }
    else
    {
        int mid = (begin+end)>>1;
        if(i<=mid)
        {
            update(begin,mid,i,j,n*2);
        }
        if(j>mid)
        {
            update(mid+1,n*2+1);
        }
        if(flag[2*n])
        {
            refresh(begin,2*n);
        }
        if(flag[2*n+1])
        {
            refresh(mid+1,2*n+1);
        }
        M[n] = M[n*2]+ M[n*2+1];
    }
}
int query(int begin,n);
    }
    if(begin>=i && end<=j)
    {
        return M[n];
    }
    if(begin>=end)
    {
        return 0;
    }
    int mid = (begin+end)>>1;
    int l=0,r=0;
    if(i<=mid)
    {
        l = query(begin,n*2);
    }
    if(j>mid)
    {
        r = query(mid+1,n*2+1);
    }
    if(flag[2*n])
    {
        refresh(begin,2*n);
    }
    if(flag[2*n+1])
    {
        refresh(mid+1,2*n+1);
    }
    M[n] = M[n*2]+ M[n*2+1];
    return l+r;
}
int main()
{
    memset(M,sizeof M);
    int n,m,a,b,c;
    scanf("%d%d",&n,&m);
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        if(a==0)
        {
            update(1,n,c);
        }
        else
        {
            printf("%dn",query(1,c));
        }
    }
    return 0;
}