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
1 #include <bits/stdc++.h> 2 #define Vector Point 3 using namespace std; 4 typedef long long ll; 5 const double eps = 1e-9; 6 const double PI=acos(-1); 7 const int maxn = 1e5+10,inf=0x3f3f3f3f; 8 //head------------------------------------------------------------------- 9 int dcmp(double x) { return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1); } 10 struct Point { 11 double x, y; 12 13 Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数 14 Point(double x = 0.0, double y = 0.0): x(x), y(y) { } //构造函数 15 16 friend istream& operator >> (istream& in, Point& P) { return in >> P.x >> P.y; } 17 friend ostream& operator << (ostream& out, const Point& P) { return out << P.x << ' ' << P.y; } 18 19 friend Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); } 20 friend Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); } 21 friend Vector operator * (const Vector& A, const double& p) { return Vector(A.x*p, A.y*p); } 22 friend Vector operator / (const Vector& A, const double& p) { return Vector(A.x/p, A.y/p); } 23 friend bool operator == (const Point& A, const Point& B) { return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0; } 24 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); } 25 void in() { scanf("%lf%lf", &x, &y); } 26 void out() { printf("%.10f %.10f\n", x, y); } 27 }; 28 29 template <class T> T sqr(T x) { return x * x;} 30 double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; } //点积 31 double Length(const Vector& A){ return sqrt(Dot(A, A)); } //向量长度 32 double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B)/Length(A)/Length(B)); } //AB向量夹角 33 double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; } //AB叉积 有向面积 34 double Area(const Point& A, const Point& B, const Point& C) { return fabs(Cross(B-A, C-A)); } 35 //向量旋转 rad正表示逆时针 反之顺时针 36 Vector Rotate(Vector A,double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));} 37 //A不能是0向量 38 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 39 double angle(Vector A) { return atan2(A.y, A.x);} //向量极角 向量A与x轴的夹角 40 Vector vecunit(Vector A){ return A / Length(A);} //单位向量 41 42 struct Line { 43 Point P; //直线上一点 44 Vector dir; //方向向量(半平面交中该向量左侧表示相应的半平面) 45 double ang; //极角，即从x正半轴旋转到向量dir所需要的角（弧度） 46 47 Line() { } //构造函数 48 Line(const Line& L): P(L.P), dir(L.dir), ang(L.ang) { } 49 Line(const Point& P, const Vector& dir): P(P), dir(dir) { ang = atan2(dir.y, dir.x); } 50 51 bool operator < (const Line& L) const { //极角排序 52 return ang < L.ang; 53 } 54 Point point(double t) { return P + dir*t; } 55 }; 56 57 58 //直线交点1 P+tv Q+tw 59 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) 60 { 61 Vector u=P-Q; 62 double t = Cross(w,u)/Cross(v,w); 63 return P+v*t; 64 } 65 //直线交点2 66 Point GetIntersection(Line a, Line b) 67 { 68 Vector u = a.P-b.P; 69 double t = Cross(b.dir, u) / Cross(a.dir, b.dir); 70 return a.P + a.dir*t; 71 } 72 bool SegmentProperInntersection(Point a1,Point a2,Point b1,Point b2)//线段相交判定 73 { 74 double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1), 75 c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); 76 return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 77 } 78 double DistanceToSegment(Point P,Point A,Point B) //点P到线段AB的距离 79 { 80 if(A==B) return Length(P-A); 81 Vector v1 = B - A,v2 = P - A,v3 = P - B; 82 if(dcmp(Dot(v1,v2))<0) return Length(v2); 83 else if(dcmp(Dot(v1,v3))>0) return Length(v3); 84 else return fabs(Cross(v1,v2))/Length(v1); 85 } 86 Point GetLineProjection(Point P,Point A,Point B) //点P在直线AB上的投影 87 { 88 Vector v = B-A; 89 return A+v*(Dot(v,P-A)/Dot(v,v)); 90 } 91 bool OnSegment(Point p, Point a1, Point a2) //判断点是否在线段a1a2上 不含端点 92 { 93 return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; 94 } 95 96 struct Circle { 97 Point c; //圆心 98 double r; //半径 99 100 Circle() { } 101 Circle(const Circle& rhs): c(rhs.c), r(rhs.r) { } 102 Circle(const Point& c, const double& r): c(c), r(r) { } 103 104 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 105 double area(void) const { return PI * r * r; } 106 }; 107 bool InCircle(Point x, Circle c) 108 { 109 return dcmp(c.r*c.r - Length(c.c - x)*Length(c.c - x)) >= 0; 110 } 111 //直线与圆的交点 112 int getLineCircleIntersection(Line L, Circle C, Point* sol) //函数返回交点个数 sol数组存放交点 113 { 114 Vector nor = normal(L.dir); 115 Line pl = Line(C.c, nor); 116 Point ip = GetIntersection(pl, L); 117 double dis = Length(ip - C.c); 118 if (dcmp(dis - C.r) > 0) return 0; 119 Point dxy = vecunit(L.dir) * sqrt(C.r*C.r - dis*dis); 120 int ret = 0; 121 sol[ret] = ip + dxy; 122 if (OnSegment(sol[ret], L. P, L.point(1))) ret++; 123 sol[ret] = ip - dxy; 124 if (OnSegment(sol[ret], L.P, L.point(1))) ret++; 125 return ret; 126 } 127 //圆与圆的交点 128 int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol)//函数返回交点个数 sol数组存放交点 129 { 130 double d=Length(C1.c-C2.c); 131 if(dcmp(d)==0){ 132 if(dcmp(C1.r-C2.r)==0) return -1;//两圆重合 133 return 0; 134 } 135 if(dcmp(C1.r+C2.r-d)<0) return 0; 136 if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; 137 138 double a=angle(C2.c-C1.c); 139 double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); 140 Point p1=C1.point(a-da),p2=C1.point(a+da); 141 sol.push_back(p1); 142 if(p1==p2) return 1; 143 sol.push_back(p2); 144 return 2; 145 } 146 //过点p到圆c的切线 147 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 148 { 149 Vector u=C.c-p; 150 double dist=Length(u); 151 if(dist<C.r) return 0; //点在圆内 152 else if(dcmp(dist-C.r)==0){//p在圆c上，只有一条切线 153 v[0]=Rotate(u,PI/2); 154 return 1; 155 }else{ 156 double ang = asin(C.r/dist); 157 v[0]=Rotate(u,-ang); 158 v[1]=Rotate(u,ang); 159 return 2; 160 } 161 } 162 //两圆的公切线 163 int getTangents(Circle A,Circle B,Point* a,Point* b) 164 { 165 int cnt=0; 166 if(A.r<B.r){swap(A,B);swap(a,b);} 167 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); 168 double rdiff=A.r - B.r; 169 double rsum=A.r + B.r; 170 if(d2<rdiff*rdiff) return 0;//内含 171 172 double base = atan2(B.c.y-A.c.y,B.c.x-A.c.x); 173 if(dcmp(d2)==0&&dcmp(A.r-B.r)==0) return -1;//两圆重合,无数条切线 174 if(dcmp(d2-rdiff*rdiff)==0){ //内切,1条切线 175 a[cnt]=A.point(base);b[cnt]=B.point(base);cnt++; 176 return 1; 177 } 178 //有外公切线 179 double ang=acos((A.r-B.r)/sqrt(d2)); 180 a[cnt] = A.point(base+ang);b[cnt] = B.point(base+ang);cnt++; 181 a[cnt] = A.point(base-ang);b[cnt] = B.point(base-ang);cnt++; 182 if(dcmp(d2-rsum*rsum)==0){ //一条内公切线 183 a[cnt]=b[cnt]=A.point(base);cnt++; 184 } 185 else if(dcmp(d2-rsum*rsum)>0){ //两条内公切线 186 ang = acos((A.r+B.r)/sqrt(d2)); 187 a[cnt] = A.point(base+ang);b[cnt] = B.point(PI+base+ang);cnt++; 188 a[cnt] = A.point(base-ang);b[cnt] = B.point(PI+base-ang);cnt++; 189 } 190 return cnt; 191 } 192 double SegCircleArea(Circle C, Point a, Point b) //线段切割圆 193 { 194 double a1 = angle(a - C.c); 195 double a2 = angle(b - C.c); 196 double da = fabs(a1 - a2); 197 if (da > PI) da = PI * 2.0 - da; 198 return dcmp(Cross(b - C.c, a - C.c)) * da * sqr(C.r) / 2.0; 199 } 200 201 double PolyCiclrArea(Circle C, Point *p, int n)//多边形与圆相交面积 202 { 203 double ret = 0.0; 204 Point sol[2]; 205 p[n] = p[0]; 206 for(int i=0;i<n;i++) 207 { 208 double t1, t2; 209 int cnt = getLineCircleIntersection(Line(p[i], p[i+1]-p[i]), C, sol); 210 if (cnt == 0) 211 { 212 if (!InCircle(p[i], C) || !InCircle(p[i+1], C)) ret += SegCircleArea(C, p[i], p[i+1]); 213 else ret += Cross(p[i+1] - C.c, p[i] - C.c) / 2.0; 214 } 215 if (cnt == 1) 216 { 217 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]); 218 else ret += SegCircleArea(C, p[i], sol[0]), ret += Cross(p[i+1] - C.c, sol[0] - C.c) / 2.0; 219 } 220 if (cnt == 2) 221 { 222 if ((p[i] < p[i + 1]) ^ (sol[0] < sol[1])) swap(sol[0], sol[1]); 223 ret += SegCircleArea(C, p[i], sol[0]); 224 ret += Cross(sol[1] - C.c, sol[0] - C.c) / 2.0; 225 ret += SegCircleArea(C, sol[1], p[i+1]); 226 } 227 } 228 return fabs(ret); 229 } 230 double PolygonArea(Point *po, int n) //多边形面积 231 { 232 double area = 0.0; 233 for(int i = 1; i < n-1; i++) { 234 area += Cross(po[i]-po[0], po[i+1]-po[0]); 235 } 236 return area * 0.5; 237 } 238 //----------------------------------------------------------------------------------- 239 Point p[maxn]; 240 int main() 241 { 242 double a,w,h; 243 int n; 244 cin>>a>>w>>h>>n; 245 double ang=a/180*PI; 246 double minx=inf,maxx=-1.0; 247 double miny=inf,maxy=-1.0; 248 for(int i=0;i<n;i++) 249 { 250 p[i].in(); 251 p[i]=p[0]+Rotate(p[i]-p[0],ang); 252 minx=min(minx,p[i].x); 253 miny=min(miny,p[i].y); 254 maxx=max(maxx,p[i].x); 255 maxy=max(maxy,p[i].y); 256 } 257 double dux=w/(maxx-minx); 258 double duy=h/(maxy-miny); 259 for(int i=0;i<n;i++) 260 { 261 p[i].x-=minx; 262 p[i].y-=miny; 263 p[i].x*=dux; 264 p[i].y*=duy; 265 p[i].out(); 266 } 267 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e4+20,mod=1e9+7,inf=0x3f3f3f3f; 4 typedef long long ll; 5 struct edge 6 { 7 int from,to,c,f; 8 edge(int u,int v,int c,int f):from(u),to(v),c(c),f(f) {} 9 }; 10 int n,m; 11 vector<edge> edges; 12 vector<int> g[maxn]; 13 int d[maxn];//从起点到i的距离 14 int cur[maxn];//当前弧下标 15 void init(int N) 16 { 17 for(int i=0; i<=N; i++) g[i].clear(); 18 edges.clear(); 19 } 20 void addedge(int from,int to,int c) //加边 支持重边 21 { 22 edges.push_back(edge(from,to,c,0)); 23 edges.push_back(edge(to,from,0,0)); 24 int siz=edges.size(); 25 g[from].push_back(siz-2); 26 g[to].push_back(siz-1); 27 } 28 int bfs(int s,int t) //构造一次层次图 29 { 30 memset(d,-1,sizeof(d)); 31 queue<int> q; 32 q.push(s); 33 d[s]=0; 34 while(!q.empty()) 35 { 36 int x=q.front();q.pop(); 37 for(int i=0;i<g[x].size();i++) 38 { 39 edge &e=edges[g[x][i]]; 40 if(d[e.to]<0&&e.f<e.c) //d[e.to]=-1表示没访问过 41 { 42 d[e.to]=d[x]+1; 43 q.push(e.to); 44 } 45 } 46 } 47 return d[t]; 48 } 49 int dfs(int x,int a,int t) // a表示x点能接收的量 50 { 51 if(x==t||a==0)return a; 52 int flow=0,f;//flow总的增量 f一条增广路的增量 53 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 54 { 55 edge &e=edges[g[x][i]]; 56 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 57 { 58 e.f+=f; 59 edges[g[x][i]^1].f-=f; //修改流量 60 flow+=f; 61 a-=f; 62 if(a==0) break; 63 } 64 } 65 return flow; 66 } 67 int maxflow(int s,int t) 68 { 69 int flow=0; 70 while(bfs(s,t)!=-1) //等于-1代表构造层次图失败 结束 71 { 72 memset(cur,0,sizeof(cur)); 73 flow+=dfs(s,inf,t); 74 } 75 return flow; 76 } 77 struct node 78 { 79 ll first,second; 80 }dog[maxn],bowl[maxn]; 81 int num[maxn]; 82 ll s; 83 int main() 84 { 85 while(scanf("%d%d%lld",&n,&m,&s)!=EOF) 86 { 87 init(n+2*m+1); 88 for(int i=1;i<=n;i++) 89 scanf("%lld%lld",&dog[i].first,&dog[i].second); 90 for(int i=1;i<=m;i++) 91 scanf("%lld%lld%d",&bowl[i].first,&bowl[i].second,&num[i]); 92 for(int i=1;i<=n;i++) 93 { 94 addedge(0,i,1); 95 for(int j=1;j<=m;j++) 96 { 97 if(sqrt(pow(abs(dog[i].first-bowl[j].first),2)+pow(abs(dog[i].second-bowl[j].second),2))+10<=s) 98 addedge(i,n+j,1); 99 } 100 } 101 for(int j=1;j<=m;j++) 102 addedge(n+j,n+m+j,num[j]), 103 addedge(n+m+j,n+2*m+1,inf); 104 if(maxflow(0,n+m*2+1)==n) 105 printf("YES\n"); 106 else 107 printf("NO\n"); 108 } 109 }
