LA3905 Meteor

https://vjudge.net/problem/UVALive-3905

计算出每个点在相框中的时间段，扫描线做即可

关键在如何快速计算每个点在相框中的时间段。对每个点进行运动分解，进入的时间L是x进入和y进入的最大，退出时间R是x退出和y退出的最小，画图看一下即可

画图同时可以发现不经过相框的时候会导致L >= R

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 100000 + 10;

int tt,w,h,n,x[MAXN],y[MAXN],a[MAXN],b[MAXN],flag[MAXN << 1], cnt[MAXN << 1], tot, t[MAXN << 1], ans, tag;

bool cmp(int a, int b)
{
    return t[a] == t[b] ? flag[a] > flag[b] : t[a] < t[b];
}

inline void update(int x, int a, int w, int &L, int &R)
{
    if(a == 0)
    {
        if(x <= 0 || x >= w) R = L - 1; 
    }
    else if(a > 0)
    {
        L = max(L, -x * 2520 / a);
        R = min(R, (w - x) * 2520 / a); 
    }
    else 
    {
        L = max(L, (w - x) * 2520 / a);
        R = min(R, -x * 2520 / a); 
    }
}

int main()
{
    read(tt);
    for(;tt;--tt)
    {
        read(w), read(h), read(n);
        for(register int i = 1;i <= n;++ i) read(x[i]), read(y[i]), read(a[i]), read(b[i]);
        tot = 0;
        for(register int i = 1;i <= n;++ i)
        {
            int L = 0, R = INF;
            update(x[i], a[i], w, L, R);
            update(y[i], b[i], h, L, R);
            if(L < R)
            {
                t[++tot] = L;flag[tot] = 0;cnt[tot] = tot;
                t[++tot] = R;flag[tot] = 1;cnt[tot] = tot;
            }
        }
        std::sort(cnt + 1, cnt + 1 + tot, cmp);
        ans = tag = 0;
        for(register int i = 1;i <= tot;++ i)
        {
            if(flag[cnt[i]]) -- tag;
            else ++ tag;
            ans = max(ans, tag);
        }
        printf("%d
", ans);
    }
    return 0;
}