Gym 101917 E 简单计算几何，I 最大流

题目链接 https://codeforces.com/gym/101917

E 

题意：给定一个多边形（n个点），然后逆时针旋转A度，然后对多边形进行规约，每个点的x规约到[0,w]范围内，y规约到[0,h]范围内，输出规约后的结果。

解析：求出来 多边形的长和宽，再和w,h比较，对点按比例进行缩放就好了。 （多边形旋转其实是绕给出的第一个点旋转,以为是绕原点wa了1发）。

AC代码

#include <bits/stdc++.h>
#define Vector Point
using namespace std;
typedef long long ll;
const double eps = 1e-9;
const double PI=acos(-1);
const int maxn = 1e5+10,inf=0x3f3f3f3f;
//head-------------------------------------------------------------------
int dcmp(double x) { return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1); }
struct Point {
    double x, y;

    Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数
    Point(double x = 0.0, double y = 0.0): x(x), y(y) { }   //构造函数

    friend istream& operator >> (istream& in, Point& P) { return in >> P.x >> P.y; }
    friend ostream& operator << (ostream& out, const Point& P) { return out << P.x << ' ' << P.y; }

    friend Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); }
    friend Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); }
    friend Vector operator * (const Vector& A, const double& p) { return Vector(A.x*p, A.y*p); }
    friend Vector operator / (const Vector& A, const double& p) { return Vector(A.x/p, A.y/p); }
    friend bool operator == (const Point& A, const Point& B) { return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0; }
    friend bool operator < (const Point& A, const Point& B) { return dcmp(A.x - B.x)<0 || (dcmp(A.x - B.x)==0 && dcmp(A.y - B.y)<0); }
    void in() { scanf("%lf%lf", &x, &y); }
    void out() { printf("%.10f %.10f
", x, y); }
};

template <class T> T sqr(T x) { return x * x;}
double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; }  //点积
double Length(const Vector& A){ return sqrt(Dot(A, A)); } //向量长度
double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B)/Length(A)/Length(B)); }  //AB向量夹角
double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }    //AB叉积 有向面积
double Area(const Point& A, const Point& B, const Point& C) { return fabs(Cross(B-A, C-A)); }
//向量旋转 rad正表示逆时针 反之顺时针
Vector Rotate(Vector A,double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
//A不能是0向量
Vector normal(Vector A) { return Point(-A.y, A.x) / Length(A);} //向量A的单位法向量，即A左转90度 以后把长度归归一化
double angle(Vector A) { return atan2(A.y, A.x);} //向量极角 向量A与x轴的夹角
Vector vecunit(Vector A){ return A / Length(A);} //单位向量

struct Line {
    Point P;    //直线上一点
    Vector dir; //方向向量(半平面交中该向量左侧表示相应的半平面)
    double ang; //极角，即从x正半轴旋转到向量dir所需要的角（弧度）

    Line() { }  //构造函数
    Line(const Line& L): P(L.P), dir(L.dir), ang(L.ang) { }
    Line(const Point& P, const Vector& dir): P(P), dir(dir) { ang = atan2(dir.y, dir.x); }

    bool operator < (const Line& L) const { //极角排序
        return ang < L.ang;
    }
    Point point(double t) { return P + dir*t; }
};


//直线交点1 P+tv Q+tw
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
    Vector u=P-Q;
    double t = Cross(w,u)/Cross(v,w);
    return P+v*t;
}
//直线交点2
Point GetIntersection(Line a, Line b)
{
    Vector u = a.P-b.P;
    double t = Cross(b.dir, u) / Cross(a.dir, b.dir);
    return a.P + a.dir*t;
}
bool SegmentProperInntersection(Point a1,Point a2,Point b1,Point b2)//线段相交判定
{
    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
           c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
double DistanceToSegment(Point P,Point A,Point B) //点P到线段AB的距离
{
    if(A==B) return Length(P-A);
    Vector v1 = B - A,v2 = P - A,v3 = P - B;
    if(dcmp(Dot(v1,v2))<0) return Length(v2);
    else if(dcmp(Dot(v1,v3))>0) return Length(v3);
    else return fabs(Cross(v1,v2))/Length(v1);
}
Point GetLineProjection(Point P,Point A,Point B) //点P在直线AB上的投影
{
    Vector v = B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}
bool OnSegment(Point p, Point a1, Point a2) //判断点是否在线段a1a2上 不含端点
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

struct Circle {
    Point c;    //圆心
    double r;   //半径

    Circle() { }
    Circle(const Circle& rhs): c(rhs.c), r(rhs.r) { }
    Circle(const Point& c, const double& r): c(c), r(r) { }

    Point point(double ang) const { return Point(c.x + cos(ang)*r, c.y + sin(ang)*r); } //圆心角所对应的点
    double area(void) const { return PI * r * r; }
};
bool InCircle(Point x, Circle c)
{
    return dcmp(c.r*c.r - Length(c.c - x)*Length(c.c - x)) >= 0;
}
//直线与圆的交点
int getLineCircleIntersection(Line L, Circle C, Point* sol) //函数返回交点个数 sol数组存放交点
{
    Vector nor = normal(L.dir);
    Line pl = Line(C.c, nor);
    Point ip = GetIntersection(pl, L);
    double dis = Length(ip - C.c);
    if (dcmp(dis - C.r) > 0) return 0;
    Point dxy = vecunit(L.dir) * sqrt(C.r*C.r - dis*dis);
    int ret = 0;
    sol[ret] = ip + dxy;
    if (OnSegment(sol[ret], L. P, L.point(1))) ret++;
    sol[ret] = ip - dxy;
    if (OnSegment(sol[ret], L.P, L.point(1))) ret++;
    return ret;
}
//圆与圆的交点
int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol)//函数返回交点个数 sol数组存放交点
{
    double d=Length(C1.c-C2.c);
    if(dcmp(d)==0){
        if(dcmp(C1.r-C2.r)==0) return -1;//两圆重合
        return 0;
    }
    if(dcmp(C1.r+C2.r-d)<0) return 0;
    if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0;

    double a=angle(C2.c-C1.c);
    double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));
    Point p1=C1.point(a-da),p2=C1.point(a+da);
    sol.push_back(p1);
    if(p1==p2) return 1;
    sol.push_back(p2);
    return 2;
}
//过点p到圆c的切线
int getTangents(Point p,Circle C,Vector* v)//函数返回条数,v[i]是第i条切线的向量
{
    Vector u=C.c-p;
    double dist=Length(u);
    if(dist<C.r) return 0;  //点在圆内
    else if(dcmp(dist-C.r)==0){//p在圆c上，只有一条切线
        v[0]=Rotate(u,PI/2);
        return 1;
    }else{
        double ang = asin(C.r/dist);
        v[0]=Rotate(u,-ang);
        v[1]=Rotate(u,ang);
        return 2;
    }
}
//两圆的公切线
int getTangents(Circle A,Circle B,Point* a,Point* b)
{
    int cnt=0;
    if(A.r<B.r){swap(A,B);swap(a,b);}
    double d2=(A.c.x-B.c.x)*(A.c.x-B.c.x)+(A.c.y-B.c.y)*(A.c.y-B.c.y);
    double rdiff=A.r - B.r;
    double rsum=A.r + B.r;
    if(d2<rdiff*rdiff) return 0;//内含

    double base = atan2(B.c.y-A.c.y,B.c.x-A.c.x);
    if(dcmp(d2)==0&&dcmp(A.r-B.r)==0) return -1;//两圆重合,无数条切线
    if(dcmp(d2-rdiff*rdiff)==0){                 //内切,1条切线
        a[cnt]=A.point(base);b[cnt]=B.point(base);cnt++;
        return 1;
    }
    //有外公切线
    double ang=acos((A.r-B.r)/sqrt(d2));
    a[cnt] = A.point(base+ang);b[cnt] = B.point(base+ang);cnt++;
    a[cnt] = A.point(base-ang);b[cnt] = B.point(base-ang);cnt++;
    if(dcmp(d2-rsum*rsum)==0){       //一条内公切线
        a[cnt]=b[cnt]=A.point(base);cnt++;
    }
    else if(dcmp(d2-rsum*rsum)>0){   //两条内公切线
        ang = acos((A.r+B.r)/sqrt(d2));
        a[cnt] = A.point(base+ang);b[cnt] = B.point(PI+base+ang);cnt++;
        a[cnt] = A.point(base-ang);b[cnt] = B.point(PI+base-ang);cnt++;
    }
    return cnt;
}
double SegCircleArea(Circle C, Point a, Point b) //线段切割圆
{
    double a1 = angle(a - C.c);
    double a2 = angle(b - C.c);
    double da = fabs(a1 - a2);
    if (da > PI) da = PI * 2.0 - da;
    return dcmp(Cross(b - C.c, a - C.c)) * da * sqr(C.r) / 2.0;
}

double PolyCiclrArea(Circle C, Point *p, int n)//多边形与圆相交面积
{
    double ret = 0.0;
    Point sol[2];
    p[n] = p[0];
    for(int i=0;i<n;i++)
    {
        double t1, t2;
        int cnt = getLineCircleIntersection(Line(p[i], p[i+1]-p[i]), C, sol);
        if (cnt == 0)
        {
            if (!InCircle(p[i], C) || !InCircle(p[i+1], C)) ret += SegCircleArea(C, p[i], p[i+1]);
            else ret += Cross(p[i+1] - C.c, p[i] - C.c) / 2.0;
        }
        if (cnt == 1)
        {
            if (InCircle(p[i], C) && !InCircle(p[i+1], C)) ret += Cross(sol[0] - C.c, p[i] - C.c) / 2.0, ret += SegCircleArea(C, sol[0], p[i+1]);
            else ret += SegCircleArea(C, p[i], sol[0]), ret += Cross(p[i+1] - C.c, sol[0] - C.c) / 2.0;
        }
        if (cnt == 2)
        {
            if ((p[i] < p[i + 1]) ^ (sol[0] < sol[1])) swap(sol[0], sol[1]);
            ret += SegCircleArea(C, p[i], sol[0]);
            ret += Cross(sol[1] - C.c, sol[0] - C.c) / 2.0;
            ret += SegCircleArea(C, sol[1], p[i+1]);
        }
    }
    return fabs(ret);
}
double PolygonArea(Point *po, int n) //多边形面积
{
    double area = 0.0;
    for(int i = 1; i < n-1; i++) {
        area += Cross(po[i]-po[0], po[i+1]-po[0]);
    }
    return area * 0.5;
}
//-----------------------------------------------------------------------------------
Point p[maxn];
int main()
{
    double a,w,h;
    int n;
    cin>>a>>w>>h>>n;
    double ang=a/180*PI;
    double minx=inf,maxx=-1.0;
    double miny=inf,maxy=-1.0;
    for(int i=0;i<n;i++)
    {
        p[i].in();
        p[i]=p[0]+Rotate(p[i]-p[0],ang);
        minx=min(minx,p[i].x);
        miny=min(miny,p[i].y);
        maxx=max(maxx,p[i].x);
        maxy=max(maxy,p[i].y);
    }
    double dux=w/(maxx-minx);
    double duy=h/(maxy-miny);
    for(int i=0;i<n;i++)
    {
        p[i].x-=minx;
        p[i].y-=miny;
        p[i].x*=dux;
        p[i].y*=duy;
        p[i].out();
    }
}

I 

题意：给出n个狗的坐标，m个碗的坐标和碗里的数的数量w L(升) , 狗每一秒可以走一个单位长度，狗喝水需要10s ，问S秒内每只狗从自己的位置出发是否全部都能喝完1L水。

题解：如果距离在时限内狗和碗连边，碗拆点就可以了。源点汇点不说了，很简单。

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e4+20,mod=1e9+7,inf=0x3f3f3f3f;
typedef long long ll;
struct edge
{
    int from,to,c,f;
    edge(int u,int v,int c,int f):from(u),to(v),c(c),f(f) {}
};
int n,m;
vector<edge> edges;
vector<int> g[maxn];
int d[maxn];//从起点到i的距离
int cur[maxn];//当前弧下标
void init(int N)
{
    for(int i=0; i<=N; i++) g[i].clear();
    edges.clear();
}
void addedge(int from,int to,int c) //加边 支持重边
{
    edges.push_back(edge(from,to,c,0));
    edges.push_back(edge(to,from,0,0));
    int siz=edges.size();
    g[from].push_back(siz-2);
    g[to].push_back(siz-1);
}
int bfs(int s,int t) //构造一次层次图
{
    memset(d,-1,sizeof(d));
    queue<int> q;
    q.push(s);
    d[s]=0;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=0;i<g[x].size();i++)
        {
            edge &e=edges[g[x][i]];
            if(d[e.to]<0&&e.f<e.c) //d[e.to]=-1表示没访问过
            {
                d[e.to]=d[x]+1;
                q.push(e.to);
            }
        }
    }
    return d[t];
}
int dfs(int x,int a,int t) // a表示x点能接收的量
{
    if(x==t||a==0)return a;
    int flow=0,f;//flow总的增量 f一条增广路的增量
    for(int &i=cur[x];i<g[x].size();i++)//cur[i] &引用修改其值 从上次考虑的弧
    {
        edge &e=edges[g[x][i]];
        if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.c-e.f),t))>0)    //按照层次图增广 满足容量限制
        {
            e.f+=f;
            edges[g[x][i]^1].f-=f;  //修改流量
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    return flow;
}
int maxflow(int s,int t)
{
    int flow=0;
    while(bfs(s,t)!=-1) //等于-1代表构造层次图失败 结束
    {
        memset(cur,0,sizeof(cur));
        flow+=dfs(s,inf,t);
    }
    return flow;
}
struct node
{
    ll first,second;
}dog[maxn],bowl[maxn];
int num[maxn];
ll s;
int main()
{
    while(scanf("%d%d%lld",&n,&m,&s)!=EOF)
    {
        init(n+2*m+1);
        for(int i=1;i<=n;i++)
            scanf("%lld%lld",&dog[i].first,&dog[i].second);
        for(int i=1;i<=m;i++)
            scanf("%lld%lld%d",&bowl[i].first,&bowl[i].second,&num[i]);
        for(int i=1;i<=n;i++)
        {
            addedge(0,i,1);
            for(int j=1;j<=m;j++)
            {
                if(sqrt(pow(abs(dog[i].first-bowl[j].first),2)+pow(abs(dog[i].second-bowl[j].second),2))+10<=s)
                    addedge(i,n+j,1);
            }
        }
        for(int j=1;j<=m;j++)
            addedge(n+j,n+m+j,num[j]),
            addedge(n+m+j,n+2*m+1,inf);
        if(maxflow(0,n+m*2+1)==n)
            printf("YES
");
        else
            printf("NO
");
    }
}