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来源：互联网发布：kafka java获取offset 编辑：程序博客网时间：2024/05/22 06:07

2016 wf G oil
题意，从suface挖一个井，穿过一些油田，油田是平行于surface的一些线段，获得一些value，问value最大多少。
做法：最优的挖井方法肯定经过油田的端点，枚举端点，其他点极角排序，扫描线扫一遍，更新答案。trick：因为要从地面开始挖井，所以y轴一样的油田不用算，eps要1e-12。
这里

#include <bits/stdc++.h>using namespace std;#define LL long long#define pii pair<int, int>#define MP make_pair#define inf 0x3f3f3f3f#define mod 1000000007#define eps 1e-12#define Pi acos(-1.0)#define N 2050int dcmp(double x){    if(fabs(x) < eps) return 0;    return x < 0 ? -1 : 1;}struct point{    double x, y;    point(double x = 0, double y = 0) : x(x), y(y) {}    point operator - (const point &b) const {        return point(x - b.x, y - b.y);    }    point operator + (const point &b) const {        return point(x + b.x, y + b.y);    }};struct seg{    point u, v;    seg() {}    seg(point _u, point _v){        u = _u, v = _v;    }};struct node{    point u;    LL val;    int in;    double ang;    node(){}    node(point _u, LL _val){        u = _u, val = _val;        ang = atan2(u.y, u.x);    }    bool operator < (const node &b) const {        if(dcmp(ang - b.ang) == 0) return in > b.in;        return dcmp(ang - b.ang) < 0;    }};seg a[N];node p[N*4];int n;LL ans, val[N];void gao(point u, int id, int sta){    int tot = 0;    LL cur = sta;    for(int i = 1; i <= n; ++i){        if(i == id) continue;        if(a[i].u.y == u.y) continue;        p[tot++] = node(a[i].u - u, val[i]);        p[tot++] = node(a[i].v - u, val[i]);        if(dcmp(p[tot-2].ang - p[tot-1].ang) <= 0)            p[tot-2].in = 1, p[tot-1].in = -1;        else             p[tot-2].in = -1, p[tot-1].in = 1;    }    sort(p, p + tot);    for(int i = tot; i < 2 * tot; ++i)        p[i] = p[i-tot], p[i].ang += 2 * Pi;    LL add = 0;    for(int i = 0, ps = 0; i < tot; ++i){        double nxt = p[i].ang + Pi;        while(dcmp(p[ps].ang - nxt) <= 0){            add += p[ps].val * p[ps].in;            ps++;        }        cur += p[i].val * p[i].in;        ans = max(ans, cur + add);    }}int main(){    ans = 0;    scanf("%d", &n);    for(int i = 1; i <= n; ++i){        int x0, x1, y;        scanf("%d%d%d", &x0, &x1, &y);        a[i].u = point(x0, y);        a[i].v = point(x1, y);        val[i] = (LL)abs(x0 - x1);        ans = max(ans, val[i]);    }    for(int i = 1; i <= n; ++i){        gao(a[i].u, i, val[i]);        gao(a[i].v, i, val[i]);    }    cout << ans << endl;    return 0;}

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