BZOJ 1818 内部白点(离散化+树状数组)

此题就是1227 的弱化版。

画个图或者稍微证明一下就能够知道，一定不会超过一次变换。

那么我们只需要统计有多少个白点会变黑，换句话说就是有多少个白点上下左右都有黑点。

离散化横坐标，因为没有黑点在的列是没有任何意义的，对答案也没有贡献。

然后处理每一行，对于每一行，维护一个BIT也就是哪些点会产生贡献，这个BIT最多只会有n次修改，n次查询。

所以时间复杂度为O(nlogn).

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi 3.1415926535

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Node{int x, y;}node[N];

VI v, vv;

int col[N], vis[N], tree[N], n;

bool comp(Node a, Node b){

    if (a.y==b.y) return a.x<b.x;

    return a.y>b.y;

}

void add(int x, int val){while (x<=n) tree[x]+=val, x+=lowbit(x);}

int query(int x){

    int res=;

    while (x) res+=tree[x], x-=lowbit(x);

    return res;

}

int main ()

{

    LL ans=;

    scanf("%d",&n);

    FOR(i,,n) scanf("%d%d",&node[i].x,&node[i].y), v.pb(node[i].x);

    sort(v.begin(),v.end());

    int siz=unique(v.begin(),v.end())-v.begin();

    sort(node+,node+n+,comp);

    FOR(i,,n) {

        node[i].x=lower_bound(v.begin(),v.begin()+siz,node[i].x)-v.begin()+;

        ++col[node[i].x];

    }

    int now=;

    while (now<=n) {

        vv.clear(); vv.pb(node[now].x);

        ++now;

        while (now<=n&&node[now].y==node[now-].y) vv.pb(node[now].x), ++now;

        siz=vv.size();

        FO(i,,siz) ans+=(query(vv[i]-)-query(vv[i-]));

        FO(i,,siz) {

            ++vis[vv[i]];

            if (vis[vv[i]]==&&col[vv[i]]>) add(vv[i],);

            else if (vis[vv[i]]==col[vv[i]]&&col[vv[i]]>) add(vv[i],-);

        }

    }

    printf("%lld\n",ans+n);

    return ;

}