【LOJ】#2082. 「JSOI2016」炸弹攻击 2

题解

想到n3发现思路有点卡住了

对于每个发射塔把激光塔和敌人按照极角排序，对于一个激光塔，和它转角不超过pi的激光塔中间夹的敌人总和就是答案

记录前缀和，用two-Points扫一下就行

代码

#include <bits/stdc++.h>

#define enter putchar('\n')

#define space putchar(' ')

#define pii pair<int,int>

#define fi first

#define se second

#define mp make_pair

#define MAXN 805

#define mo 99994711

#define pb push_back

#define eps 1e-8

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef unsigned long long u64;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

        if(c == '-') f = -1;

        c = getchar();

    }

    while(c >= '0' && c <= '9') {

        res = res * 10 - '0' + c;

        c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) out(x / 10);

    putchar('0' + x % 10);

}

struct Point {

    db x,y;

    Point(){}

    Point(db _x,db _y) {x = _x;y = _y;}

    friend Point operator + (const Point &a,const Point &b) {return Point(a.x + b.x,a.y + b.y);}

    friend Point operator - (const Point &a,const Point &b) {return Point(a.x - b.x,a.y - b.y);}

    friend Point operator * (const Point &a,const db &d) {return Point(a.x * d,a.y * d);}

    friend Point operator / (const Point &a,const db &d) {return Point(a.x / d,a.y / d);}

    friend db operator * (const Point &a,const Point &b) {return a.x * b.y - a.y * b.x;}

    friend db dot(const Point &a,const Point &b) {return a.x * b.x + a.y * b.y;}

    db norm() {return sqrt(x * x + y * y);}

    friend bool operator < (const Point &a,const Point &b) {return a.x < b.x || (a.x == b.x && a.y < b.y);}

}D1[MAXN],S1[MAXN],T1[MAXN];

int D,S,T,tot;

pair<db,Point> L[MAXN * 4];

Point P[MAXN * 4];

int sum[MAXN * 4];

void Solve() {

    read(D);

    for(int i = 1 ; i <= D ; ++i) scanf("%lf%lf",&D1[i].x,&D1[i].y);

    read(S);

    for(int i = 1 ; i <= S ; ++i) scanf("%lf%lf",&S1[i].x,&S1[i].y);

    read(T);

    int64 ans = 0;

    for(int i = 1 ; i <= T ; ++i) scanf("%lf%lf",&T1[i].x,&T1[i].y);

    for(int i = 1 ; i <= S ; ++i) {

        tot = 0;

        for(int j = 1 ; j <= D ; ++j) L[++tot] = mp(atan2(D1[j].y - S1[i].y,D1[j].x - S1[i].x),D1[j]);

        for(int j = 1 ; j <= T ; ++j) L[++tot] = mp(atan2(T1[j].y - S1[i].y,T1[j].x - S1[i].x),T1[j]);

        sort(L + 1,L + tot + 1);

        for(int i = 1 ; i <= tot ; ++i) P[i] = L[i].se;

        for(int i = 1 ; i <= tot ; ++i) P[i + tot] = L[i].se;

        tot *= 2;

        for(int i = 1 ; i <= tot ; ++i) {

            sum[i] = sum[i - 1];

            if(P[i].y > 0) sum[i]++;

        }

        int r = 0,cnt = 0;

        int64 s = 0;

        for(int l = 1 ; l <= tot / 2; ++l) {

            while(r < l) {if(P[r + 1].y < 0) s += sum[r + 1],++cnt;++r;}

            while((P[l] - S1[i]) * (P[r + 1] - S1[i]) > 0) {

                if(P[r + 1].y < 0) s += sum[r + 1],++cnt;++r;

            }

            if(P[l].y < 0) {

                s -= sum[l];--cnt;

                ans += s - 1LL * cnt * sum[l];

            }

        }

    }

    out(ans);enter;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}

貌似板刷了一页LOJ了呢。。