ASC 03 题解

A Areas

给不超过80个圆，问把平面分成几份?

欧拉公式，注意不同连通块

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define ALL(a) (a).begin(), (a).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll sqr(ll a){return a*a;}ld sqr(ld a){return a*a;}double sqr(double a){return a*a;}const double eps=1e-8;int dcmp(double x) {    if (fabs(x)<eps) return 0; else return x<0 ? -1 : 1; }ld PI = 3.141592653589793238462643383;class P{public:    double x,y;    P(double x=0,double y=0):x(x),y(y){}    friend long double dis2(P A,P B){return sqr(A.x-B.x)+sqr(A.y-B.y);  }    friend long double Dot(P A,P B) {return A.x*B.x+A.y*B.y; }    friend long double Length(P A) {return sqrt(Dot(A,A)); }    friend long double Angle(P A,P B) {return acos(Dot(A,B) / Length(A) / Length(B) ); }    friend P operator- (P A,P B) { return P(A.x-B.x,A.y-B.y); }    friend P operator+ (P A,P B) { return P(A.x+B.x,A.y+B.y); }    friend P operator* (P A,double p) { return P(A.x*p,A.y*p); }    friend P operator/ (P A,double p) { return P(A.x/p,A.y/p); }    friend bool operator< (const P& a,const P& b) {return dcmp(a.x-b.x)<0 ||(dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)<0 );}}; bool operator==(const P& a,const P& b) {    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y) == 0;} typedef P V;double Cross(V A,V B) {return A.x*B.y - A.y*B.x;}double Area2(P A,P B,P C) {return Cross(B-A,C-A);}P GetLineIntersection(P p,V v,P Q,V w){    V u = p-Q;    double t = Cross(w,u)/Cross(v,w);    return p+v*t;}P GetLineIntersectionB(P p,V v,P Q,V w){    return GetLineIntersection(p,v-p,Q,w-Q);}bool SegmentProperIntersection(P a1,P a2,P b1,P b2) {     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;}bool OnSegment(P p,P a1,P a2) {    return dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p,a2-p))<=0;}P read_point() {    P a;    scanf("%lf%lf",&a.x,&a.y);    return a;   } typedef vector<P> Polygon;double PolygonArea(Polygon &p) {    double area=0;    int n=p.size();    For(i,n-2) area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;} struct Edge{    int from,to;    double ang;    Edge(int _from,int _to,double _ang):from(_from),        to(_to),ang(_ang){}};#define MAXN (11111+10)struct PSLG {    int n,m,face_cnt;    ld x[MAXN],y[MAXN];    vector<Edge> edges;    vi G[MAXN];    int vis[MAXN*2],left[MAXN*2],prev[MAXN*2];    vector<Polygon> faces;    double area[MAXN];    void init(int n) {        this->n=n;        Rep(i,n) G[i].clear();        edges.clear();        faces.clear();    }    double getAngle(int from,int to) {        return atan2(y[to]-y[from],x[to]-x[from]);    }    void AddEdge(int from,int to) {        edges.pb(Edge(from,to,getAngle(from,to)));        edges.pb(Edge(to,from,getAngle(to,from)));        m=SI(edges);        G[from].pb(m-2);        G[to].pb(m-1);    }    void Build() {         Rep(u,n) {            int d=SI(G[u]);            Rep(i,d) {                Fork(j,i+1,d-1) {                    if (edges[G[u][i]].ang>edges[G[u][j]].ang) {                        swap(G[u][i],G[u][j]);                    }                }            }            Rep(i,d) {                prev[G[u][(i+1)%d]]=G[u][i];            }        }        MEM(vis)        face_cnt=0;        Rep(u,n) {            Rep(i,SI(G[u])) {                int e=G[u][i];                if (!vis[e]) {                    face_cnt++;                    Polygon poly;                    while(1) {                        vis[e]=1;                        left[e]=face_cnt;                        int from = edges[e].from;                        P p(x[from],y[from]);                        poly.pb(p);                        e=prev[e^1];                        if (e==G[u][i]) break;                    }                    faces.pb(poly);                }            }         }        Rep(i,face_cnt) {            area[i]=PolygonArea(faces[i]);        }    }}g;int n, sz, m;int L,W;P p1[MAXN],p2[MAXN];void find_path() {    vector<P> v;    Rep(i,n) {        Fork(j,i+1,n-1)            if (dcmp(Cross(p2[i]-p1[i],p2[j]-p1[j])!=0)) {                P p=GetLineIntersectionB(p1[i],p2[i],p1[j],p2[j]);                v.pb(p);            }    }    sort(ALL(v));    v.erase( unique(ALL(v)), v.end() );    sz=SI(v);//    Rep(i,sz) cout<<v[i].x<<' '<<v[i].y<<endl;    g.init(sz);    Rep(i,sz) g.x[i]=v[i].x,g.y[i]=v[i].y;    Rep(i,n) {        vector<P> ve;        Rep(j,n) if (i!=j) {            if (dcmp(Cross(p2[i]-p1[i],p2[j]-p1[j])!=0)) {                P p=GetLineIntersectionB(p1[i],p2[i],p1[j],p2[j]);                ve.pb(p);            }        }        sort(ALL(ve));        ve.erase( unique(ALL(ve)), ve.end() );        vector<int> vid;        Rep(j,SI(ve)) {            Rep(k,sz) {                if (v[k]==ve[j]) {                    vid.pb(k);break;                }            }        }//      Rep(j,SI(ve)) cout<<vid[j]<<' ';puts("");        for(int j=1;j<SI(ve);j++) if (vid[j]!=vid[j-1]){            g.AddEdge(vid[j],vid[j-1]);        }    }    g.Build();}int isPointInPolygon(P p,Polygon poly) {    int wn=0;    int n=poly.size();    Rep(i,n) {        if (OnSegment(p,poly[i],poly[(i+1)%n])) return -1;        int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));        int d1 = dcmp(poly[i].y-p.y);        int d2 = dcmp(poly[(i+1)%n].y-p.y);        if ( k > 0 && d1 <= 0 && d2 > 0 ) wn++;        if ( k < 0 && d2 <= 0 && d1 > 0 ) wn--;    }    if (wn!=0) return 1;    return 0;}P GetLineProjection(P p,P A,P B) {    V v=B-A;    return A+v*(Dot(v,p-A)/Dot(v,v));}bool is_intesect(Polygon poly,P o,double r) {    int n=SI(poly);    if (isPointInPolygon(o,poly)==1) return 1;    Rep(i,n) {        if (dcmp(Length(poly[i]-o)-r)<0) return 1;    }    Rep(i,n) {        P A=poly[i],B=poly[(i+1)%n];        if (dcmp(Length((A+B)/2-o)-r)<0) return 1;    }    Rep(i,n) {        P A=poly[i],B=poly[(i+1)%n];        P C=GetLineProjection(o,A,B);        if (C==A||C==B) continue;        if (dcmp(Length(C-o)-r)<0 && OnSegment(C,A,B)) return 1;    }    return 0;}int main(){    freopen("areas.in","r",stdin);    freopen("areas.out","w",stdout);    n=read();    Rep(i,n) {        p1[i]=read_point(); p2[i]=read_point();    }    find_path();    vector<double > ans;    Rep(i,g.face_cnt) if (dcmp(g.area[i])>0) {        ans.pb(g.area[i]);    }    sort(ALL(ans));    cout<<SI(ans)<<endl;    Rep(i,SI(ans)) printf("%.4lf\n",ans[i]);    return 0;}

B Beloved Sons

国王要给王子们安排婚事，每个王子都有一个权重Ai，目标是使所有与心爱女孩结婚的王子们的权重的2范数最大。

二分图最大权匹配，km裸题

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll costA[410]; namespace KM{    const int N=405;    bool b[410][410];    const ll inf=1e15;    ll n,nl,nr,m,z,py,x,y,i,j,p,lk[N],pre[N];    bool vy[N];    ll lx[N],ly[N],d,w[N][N],slk[N];ll ans;    ll work(int nl,int nr){ // nl nr w      n=max(nl,nr);//      For(i,nl) For(j,nr) w[i][j]=max(0,f[i][j]);    For(i,nl) For(j,nr) w[i][j]=0;    MEM(b)    For(x,nl) {        int m=read(),y;        while(m--)            scanf("%d",&y),w[y][x]=max(w[y][x],costA[x]),b[x][y]=1;    }//    For(i,nl) {//      For(j,nr) cout<<w[i][j]<<' ';cout<<endl;//  }      for(i=1;i<=n;i++)for(j=1;j<=n;j++)lx[i]=max(lx[i],w[i][j]);      for(i=1;i<=n;i++){        for(j=1;j<=n;j++)slk[j]=inf,vy[j]=0;        for(lk[py=0]=i;lk[py];py=p){          vy[py]=1;d=inf;x=lk[py];          for(y=1;y<=n;y++)if(!vy[y]){            if(lx[x]+ly[y]-w[x][y]<slk[y])slk[y]=lx[x]+ly[y]-w[x][y],pre[y]=py;            if(slk[y]<d)d=slk[y],p=y;          }          for(y=0;y<=n;y++)if(vy[y])lx[lk[y]]-=d,ly[y]+=d;else slk[y]-=d;        }        for(;py;py=pre[py])lk[py]=lk[pre[py]];      }      for(i=1;i<=n;i++)ans+=lx[i]+ly[i];//    printf("%lld\n",ans);    for(i=1;i<=nl;i++){        printf("%I64d ",w[lk[i]][i]?((b[i][lk[i]])?lk[i]:0):0);    }    }}int n,m;int main(){//  freopen("b.in","r",stdin);    freopen("beloved.in","r",stdin);    freopen("beloved.out","w",stdout);    int n=read();    For(i,n) costA[i]=read();    For(i,n) costA[i]=costA[i]*costA[i];    KM::work(n,n);    return 0;}

D Data Transmission

求一个分层图的阻塞流(blocking flow)。(n = 1.5k; m = 0.3M)

HLLP网络流，裸交能过。

E Strong Defence

一个有向图，要给一些边染色，使得所用的颜色最多，且S到T的任意路径的都包含所有颜色。

答案下界是点S到点T的最短路长度。
构造，我们把点按到S的距离分层，每一次保证把第i层和第i+1层割开，显然成立.

F Weird Dissimilarity

字符c1和c2的距离为d(c1, c2)，已知两个字符串s和t，现在要找长度相等的两个字符串a和b，使得s是a的子序列，t是b的子序列，且a和b的距离最小。

dp[i][j]表示a的前i个字母和b的前j个字母匹配的最优代价。
然后dp做法显然。
注意读入的字符为>32的可见字符，如果用char读入，char的范围是-128~+127，hash时要注意。

G PL/Cool

为PL/Cool写一个解释器，它有两种操作：define op1 op2把以后所有的op1都用op2替换，替换是可以一直迭代的，但重复define和产生环的define应该被无视；print expr，其中expr是个四则运算表达式，输出先根据define替换再表达求值的结果。

不敢开……

I Two Cylinders

求轴垂直相交的两个高度无限圆柱的相交区域体积。

直接辛普森积分

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } typedef long long ll;typedef unsigned long long ull;typedef long double ld;int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} double r1,r2;ld PI = 3.141592653589793238462643383;namespace Simpson{    double a;    double f(double k) {        double lamda=sqrt(r2*r2-k*k);        if (lamda>=r1) return PI*r1*r1;        if (fabs(lamda)<1e-8) return 0;        double xita=acos(lamda/r1);        double S1=sqrt(r1*r1-lamda*lamda)*lamda*2;        double S2=(PI-2*xita)*r1*r1;//      cout<<S1<<' '<<S2<<' '<<S1+S2<<endl;        return S1+S2;    }    double simpson(double a,double b) {        double c=(a+b)/2;        return (f(a)+f(b)+4*f(c)) * (b-a) / 6;    }    double asr(double a,double b,double eps,double A) {        double c=(a+b)/2;        double l=simpson(a,c),r=simpson(c,b);        if (fabs(l+r-A)<=15*eps) return l+r+(l+r-A)/15.;        return asr(a,c,eps,l)+asr(c,b,eps,r);    }    double asr(double a,double b,double eps) {        return asr(a,b,eps,simpson(a,b));    }}int main(){    freopen("twocyl.in","r",stdin);    freopen("twocyl.out","w",stdout);    cin>>r1>>r2;    if (r1>r2) swap(r1,r2);    double s=PI*r1*r1*r2;    printf("%.4lf\n",Simpson::asr(0,r2,1e-8)*2);    return 0;}