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题意

分析

代码

#include <bits/stdc++.h>

using LL = long long;

const int MAXN = 1e5 + 5;

int n, m, D;

std::pair<LL, LL> p1[MAXN], p2[MAXN];

struct Event {
    LL x, y; int val;
    
    Event() = default;
    Event(LL _x, LL _y, int _v): x(_x), y(_y), val(_v) {}
}; std::vector<Event> events; // 扫描线事件 
std::vector<LL> xlist, sorted; // x离散化列表，y离散化列表 

namespace BIT {
    int sum[MAXN * 3];
    
    void Init() {
        memset(sum, 0, sizeof sum);
    }
    
#define lowbit(x) ((x) & -(x))
    
    void Add(int a, int x) {
        for (int i = a; i <= (int)sorted.size(); i += lowbit(i))
            sum[i] += x;
    }
    
    int Query(int a) {
        int res = 0;
        for (int i = a; i > 0; i -= lowbit(i))
            res += sum[i];
        return res;
    }
}

LL minAns = 0, maxAns = 0, minus = 0; // 加、减的总和 

int covered[MAXN];

int Y(LL y) {
    return std::lower_bound(sorted.begin(), sorted.end(), y) - sorted.begin() + 1;
}

LL C2(LL x) {
    return x * (x - 1) / 2;
}

void Solve() {
    events.clear(); // p2中x起止&y&+-1 
    xlist.clear(); // p2中x起止, p1中x 
    sorted.clear(); // p2中y，p1中y起止 
    
    for (int i = 1; i <= m; ++i) {
        events.emplace_back(p2[i].first - D, p2[i].second, 1);
        events.emplace_back(p2[i].first + D + 1, p2[i].second, -1);
        
        xlist.push_back(p2[i].first - D);
        xlist.push_back(p2[i].first + D + 1);
        
        sorted.push_back(p2[i].second);
    }
    for (int i = 1; i <= n; ++i) {
        xlist.push_back(p1[i].first);
        
        sorted.push_back(p1[i].second - D - 1);
        sorted.push_back(p1[i].second + D);
    }
    
    std::sort(xlist.begin(), xlist.end());
    xlist.erase(std::unique(xlist.begin(), xlist.end()), xlist.end());
    
    std::sort(sorted.begin(), sorted.end());
    sorted.erase(std::unique(sorted.begin(), sorted.end()), sorted.end());
    
    std::sort(events.begin(), events.end(), [] (const Event &a, const Event &b) { return a.x < b.x; });
    
    BIT::Init();
     
    int j1 = 0, j2 = 1;
    
    for (LL x: xlist) {
        while (j1 < (int)events.size() && events[j1].x == x) { // 把离散化的纵坐标加进去 
            BIT::Add(Y(events[j1].y), events[j1].val);
            ++j1;
        }
        
        while (j2 <= n && p1[j2].first == x) {
            covered[j2] = BIT::Query(Y(p1[j2].second + D)) - BIT::Query(Y(p1[j2].second - D - 1)); // 询问离散化的纵坐标的区间和 
            ++j2;
        }
    }
    
    std::sort(covered + 1, covered + n + 1);
    
    for (int i = 1; i <= n; ++i) {
        minAns += 1LL * (n - i) * covered[i]; // n-i个比它大 
        maxAns += 1LL * (n - i) * covered[n - i + 1]; // n-i个比它大 
        minus += C2(covered[i]); // 加到减的总和里面 
    }
}

int main() {
#ifndef LOCAL
    freopen("mhw.in", "r", stdin);
    freopen("mhw.out", "w", stdout);
#endif
    
    scanf("%d%d%d", &n, &m, &D);
    for (int i = 1; i <= n; ++i) {
        static int x, y;
        
        scanf("%d%d", &x, &y);
        p1[i] = std::make_pair(x + y, x - y);
    }
    for (int i = 1; i <= m; ++i) {
        static int x, y;
        
        scanf("%d%d", &x, &y);
        p2[i] = std::make_pair(x + y, x - y);
    }
    std::sort(p1 + 1, p1 + n + 1);
    std::sort(p2 + 1, p2 + m + 1);
    
    Solve();
    
    std::swap(n, m);
    std::swap(p1, p2);
    
    Solve();
    
    printf("%lld %lld\n", minAns - minus, maxAns - minus);
    
    return 0;
}

test20180922 世界第一的猛汉王

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原文地址：https://www.cnblogs.com/autoint/p/9710515.html