省队集训 Day3 杨北大

【题目大意】

给出平面上$n$个点$(x_i, y_i)$，请选择一个不在这$n$个点之内的点$(X, Y)$，定义$(X, Y)$的价值为往上下左右四个方向射出去直线，经过$n$个点中的数量的最小值。

Task 1: 求价值最大的点

Task 2: 求价值最大的点的个数

保证Task 1和Task 2各占50pts。

对于30%的数据，$n \leq 200$；

对于60%的数据，$n \leq 5000$；

对于100%的数据，$n \leq 300000$。

每档数据中，50%保证$1 \leq x_i, y_i \leq n$；50%保证$1 \leq x_i, y_i \leq 10^9$。

【题解】

考场写了60分暴力。。签到题不会打系列。

回家路上认真想了想，离散后把每个位置用vector存起来。

二分答案$x$，那么得到的就是若干合法区间（有横的也有竖的），那么一个点只要被一横一竖两个区间同时经过，就有1的贡献，先不考虑$(X, Y)$在$n$个点中的情况，那么这样就能用扫描线+BIT解决了。

预处理出来对于关键点的贡献，统计的时候减去即可。

复杂度$O(nlog^2n)$。

Orz YangPKU

# include <vector>
# include <stdio.h>
# include <assert.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 3e5 + 10;
const int mod = 1e9+7;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
    return x;
}

int n, x[M], y[M], X, Y, ori[M];
vector<int> psa, psb;
vector<int> vx[M], vy[M]; int xlen[M], ylen[M];

struct option {
    int op, x, yl, yr, del;
    option() {}
    option(int op, int x, int yl, int yr, int del = 0) : op(op), x(x), yl(yl), yr(yr), del(del) {}
    friend bool operator < (option a, option b) {
        return a.x < b.x || (a.x == b.x && a.op < b.op);
    }
}o[M * 3]; int on;

struct BIT {
    # define lb(x) (x&(-x))
    ll c[M]; int n;
    inline void set(int _n) {
        n = _n; memset(c, 0, sizeof c);    
    }
    inline void edt(int x, int d) {
        for (; x<=n; x+=lb(x)) c[x] += d;
    }
    inline ll sum(int x) {
        ll ret = 0;
        for (; x; x-=lb(x)) ret += c[x];
        return ret;
    }
    inline ll sum(int x, int y) {
        if(x > y) return 0;
        return sum(y) - sum(x-1);
    }
    # undef lb
}T;

inline ll chk(int x) {
    ll ans = 0;
    on = 0;
    for (int i=1, l, r; i<=X; ++i) {
        if(xlen[i] >= x + x) {
            l = vx[i][x-1] + 1, r = vx[i][xlen[i]-x] - 1;  
            o[++on] = option(2, i, l, r);    
        }
    }
    for (int i=1, l, r; i<=Y; ++i) {
        if(ylen[i] >= x + x) {
            l = vy[i][x-1] + 1, r = vy[i][ylen[i]-x] - 1;
            o[++on] = option(1, l, i, i, 1);
            o[++on] = option(1, r+1, i, i, -1);
        }
    }
    sort(o+1, o+on+1);
    for (int i=1; i<=on; ++i) {
        if(o[i].op == 1) T.edt(o[i].yl, o[i].del);
        else ans += T.sum(o[i].yl, o[i].yr);
    }
    for (int i=1; i<=n; ++i) ans -= (ori[i] >= x);
    return ans;
}

int main() {
    int type;
    n = getint(); type = getint();
    for (int i=1; i<=n; ++i) psa.push_back(x[i] = getint()), psb.push_back(y[i] = getint());
    sort(psa.begin(), psa.end()); psa.erase(unique(psa.begin(), psa.end()), psa.end()); X = psa.size();
    sort(psb.begin(), psb.end()); psb.erase(unique(psb.begin(), psb.end()), psb.end()); Y = psb.size();
    for (int i=1; i<=n; ++i) x[i] = lower_bound(psa.begin(), psa.end(), x[i]) - psa.begin() + 1, y[i] = lower_bound(psb.begin(), psb.end(), y[i]) - psb.begin() + 1;
    for (int i=1; i<=n; ++i) vx[x[i]].push_back(y[i]), vy[y[i]].push_back(x[i]);
    for (int i=1; i<=X; ++i) {
        sort(vx[i].begin(), vx[i].end());
        xlen[i] = vx[i].size();
    }
    for (int i=1; i<=Y; ++i) {
        sort(vy[i].begin(), vy[i].end());
        ylen[i] = vy[i].size();
    }
    for (int i=1, xid, yid; i<=n; ++i) {
        xid = lower_bound(vy[y[i]].begin(), vy[y[i]].end(), x[i]) - vy[y[i]].begin();
        yid = lower_bound(vx[x[i]].begin(), vx[x[i]].end(), y[i]) - vx[x[i]].begin();
        ori[i] = min(min(xid, ylen[y[i]]-xid-1), min(yid, xlen[x[i]]-yid-1));
    }
    int l = 0, r = n/4, mid, ans = -1; T.set(Y); ll t;
    while(1) {
        if(r - l <= 3) {
            for (int i=r; i>=l; --i) {
                if(t = chk(i)) {
                    ans = i;
                    break;
                }
            }
            break;
        }
        mid = l+r>>1;
        if(chk(mid)) l = mid;
        else r = mid;
    }
    assert(ans != -1);
    if(type == 1) printf("%d\n", ans);
    else printf("%lld\n", t);
    return 0;
}