【Luogu】P2154虔诚的墓主人（树状数组）

　　题目链接

　　这题就是考虑我们有这样一个情况

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" alt="" width="188" height="165" />

　　然后我们试图很方便地求出第三行第二个和第三个常青树之间所有点上下常青树的组合。

　　考虑使用树状数组，一开始把数组平放在最底下，然后往上推。

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cctype>
#include<cstring>
#define maxn 200020
#define maxd 12
#define mod 2147483648
using namespace std;
inline long long read(){
    long long num=,f=;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*+ch-'';
        ch=getchar();
    }
    return num*f;
}

struct Tree{
    long long x,y;
    bool operator <(const Tree a)const{
        if(x!=a.x)    return x<a.x;
        return y<a.y;
    }
}q[maxn];

long long s[maxn*],cnt;
long long c[maxn][maxd];

long long tree[maxn*];
inline long long low(long long x){    return x&(-x);    }

inline void add(long long pos,long long lim,long long num){
    while(pos<=lim){
        tree[pos]+=num;
        tree[pos]%=mod;
        pos+=low(pos);
    }
    return;
}

inline long long query(long long pos){
    long long ans=;
    while(pos){
        ans+=tree[pos];
        ans%=mod;
        pos-=low(pos);
    }
    return ans;
}

long long up[maxn],down[maxn];
long long lef[maxn],rig[maxn];

int main(){
    long long m=read()+,n=read()+;
    long long e=read();
    for(long long i=;i<=e;++i){
        q[i]=(Tree){read(),read()};
        swap(q[i].x,q[i].y);
        s[++cnt]=q[i].x;    s[++cnt]=q[i].y;
    }
    sort(s+,s+cnt+);
    long long size=unique(s+,s+cnt+)-s-;
    for(long long i=;i<=e;++i){
        q[i].x=lower_bound(s+,s+size+,q[i].x)-s;
        q[i].y=lower_bound(s+,s+size+,q[i].y)-s;
    }
    sort(q+,q+e+);
    for(long long i=;i<=e;++i){
        up[q[i].y]++;
        rig[q[i].x]++;
    }
    long long d=read();
    for(long long i=;i<=e;++i)    c[i][]=;
    for(long long i=;i<=e;++i)
        for(long long j=;j<=d;++j)    c[i][j]=(c[i-][j]+c[i-][j-])%mod;
    long long ans=;
    for(long long i=;i<=e;++i){
        if(q[i].x==q[i-].x&&i!=){
            //printf("i=%d\nxi=%d yi=%d\n%d %d %d %d\n",i,q[i].x,q[i].y,lef[q[i].x],rig[q[i].x],query(q[i].y),query(q[i].y-q[i-1].y));
            ans+=c[lef[q[i].x]][d]*c[rig[q[i].x]][d]*(query(q[i].y-)-query(q[i-].y))%mod;
        }

        up[q[i].y]--;    down[q[i].y]++;
        lef[q[i].x]++;    rig[q[i].x]--;
        add(q[i].y,size,c[up[q[i].y]][d]*c[down[q[i].y]][d]-query(q[i].y)+query(q[i].y-));
    }
    printf("%lld\n",(ans%mod+mod)%mod);
    return ;
}