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把站位当作抵消的部分

建立两棵线段树,统计x轴和y轴被覆盖的情况

每次的答案就是

放过的行数×行长度+放过的列数×列长度-抵消块数

抵消块数就是x轴被覆盖行数\times y轴被覆盖行数 \times 2

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100011
#define ls rt<<1
#define rs rt<<1|1
int P[N<<2],Q[N<<2],n,m;
void upd(int *s,int x,int l,int r,int rt){
    if(l==r){
        s[rt]^=1;
        return;
    }
    int m=(l+r)>>1;
    if(x<=m)upd(s,x,l,m,ls);
    else upd(s,x,m+1,r,rs);
    s[rt]=s[ls]+s[rs];
}
int ask(int *s,int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R)return s[rt];
    int m=(l+r)>>1,ans=0;
    if(L<=m)ans=ask(s,L,R,l,m,ls);
    if(R>m)ans+=ask(s,L,R,m+1,r,rs);
    return ans;
}
int main(){
    fin("in");
    int q;
    in>>n>>m>>q;
    int u,x,y,X,Y;
    while(q--){
        in>>u>>x>>y;
        if(u==2){
            in>>X>>Y;
            int a=ask(P,x,X,1,n,1),b=ask(Q,y,Y,1,m,1);
            out<<(1ll*a*(Y-y+1)+1ll*b*(X-x+1)-2ll*a*b)<<ln;
        }
        else upd(P,x,1,n,1),upd(Q,y,1,m,1);
    }
}
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