【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; }
时间: 2024-11-08 19:07:45