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 }
原文地址：https://www.cnblogs.com/stranger-/p/10840551.html
时间： 2024-10-09 19:01:14