Codeforces 1139F（树状数组+扫描线）

题目传送

做法

• 对于每个人，inc为x，pref为y；对于每道菜，p和s为x，b为y
• 于是根据题意有\[p[i]<=x<=s[i]\]\[p[i]+b[i]<=x+y\]\[p[i]-b[i]<=x-y\]
• 把所有出现的点都离散化一下，然后开始扫x轴
• 对于x+y和x-y这两个函数，分别开一个树状数组去维护合法点的个数

#include <cstdio>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <vector>
#define rep(i, a, b)    for (int i = a; i <= b; i++)
#define All(x)  (x.begin()), (x.end())
using namespace std;

template <typename T> void read(T &x) {
    x = 0;
    T s = 1, c = getchar();
    for (; !isdigit(c); c = getchar())
        if (c == '-')   s = -1;
    for (; isdigit(c); c = getchar())
        x = x * 10 + c - 48;
    x *= s;
}

typedef long long ll;
const int maxn = 1e5 + 5;
int n, m, tot;
ll p[maxn], s[maxn], b[maxn], inc[maxn], pref[maxn];
int ans[maxn];
vector<ll> A, B;
vector<pair<ll, int>>que;

class FenwickTree {
public:
    int F[maxn << 1];

    void Modify(int x, int val) {
        for (; x <= maxn << 1; x += x&-x)
            F[x] += val;
    }

    int Query(int x) {
        int ret = 0;
        for (; x; x -= x&-x)
            ret += F[x];
        return ret;
    }
}bitA, bitB;

inline void Read() {
    read(n), read(m);
    rep(i, 1, n)    read(p[i]);
    rep(i, 1, n)    read(s[i]);
    rep(i, 1, n)    read(b[i]);
    rep(i, 1, m)    read(inc[i]);
    rep(i, 1, m)    read(pref[i]);
}

inline void Discrete() {
    A.push_back(-(1LL << 60));
    B.push_back(-(1LL << 60));
    rep(i, 1, n) {
        que.push_back({p[i], i});
        que.push_back({s[i], i + maxn * 2});
        A.push_back(p[i] + b[i]);
        B.push_back(p[i] - b[i]);
    }
    rep(i, 1, m) {
        que.push_back({inc[i], i + maxn});
        A.push_back(inc[i] + pref[i]);
        B.push_back(inc[i] - pref[i]);
    }

    sort(All(que)), sort(All(A)), sort(All(B));
    A.erase(unique(All(A)), A.end());
    B.erase(unique(All(B)), B.end());
}

inline void Sweep() {
    int num = 0;
    for (auto i : que) {
        int x = i.second, y;
        if (x < maxn) {
            num++;
            y = lower_bound(All(A), p[x] + b[x]) - A.begin();
            bitA.Modify(y, 1);
            y = lower_bound(All(B), p[x] - b[x]) - B.begin();
            bitB.Modify(y, 1);
        } else if (x < maxn * 2) {//注意排序顺序p < inc < s
            x -= maxn;
            y = lower_bound(All(A), inc[x] + pref[x]) - A.begin();
            ans[x] += bitA.Query(y);
            y = lower_bound(All(B), inc[x] - pref[x]) - B.begin();
            ans[x] += bitB.Query(y);
            ans[x] -= num;
            //当一个x一定大于等于另一个x时，无论y的关系如何，这样容斥一定能得到正确答案
        } else {
            x -= maxn * 2;
            num--;
            y = lower_bound(All(A), p[x] + b[x]) - A.begin();
            bitA.Modify(y, -1);
            y = lower_bound(All(B), p[x] - b[x]) - B.begin();
            bitB.Modify(y, -1); 
        }
    }
    rep(i, 1, m)    printf("%d%c", ans[i], " \n"[i == m]);
}

int main() {
    Read();
    Discrete();
    Sweep();
    return 0;
}