[KM算法] BZOJ 2539 [Ctsc2000]丘比特的烦恼

来源：互联网发布：压缩命令 linux gzip 编辑：程序博客网时间：2024/04/29 05:19
这就很裸了
只是坑点实在多
坐标有实数
字串大小写
不可连的边设为-inf
貌似题目保证有完备匹配？
#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<cstring>#include<iostream>#include<map>#include<string>#define cl(x) memset(x,0,sizeof(x))using namespace std;const int N=35;int n,ans,w[N][N];int boy[N],lx[N],ly[N],sla[N];int S[N],T[N];inline bool match(int u){    S[u]=1;    for (int v=1;v<=n;v++){        if (T[v]) continue;        if (lx[u]+ly[v]==w[u][v]){            T[v]=1;            if (!boy[v] || match(boy[v]))                return boy[v]=u,1;        }        else            sla[v]=min(sla[v],lx[u]+ly[v]-w[u][v]);    }    return 0;}inline int KM(){    cl(boy); cl(ly);    for (int i=1;i<=n;i++)    {        lx[i]=-1<<30;        for (int j=1;j<=n;j++) lx[i]=max(lx[i],w[i][j]);    }    for (int i=1;i<=n;i++)    {        for (int j=1;j<=n;j++) sla[j]=1<<30;        for (;;)        {            cl(S); cl(T);            if (match(i)) break;            int a=1<<30;            for (int j=1;j<=n;j++) if (!T[j]) a=min(a,sla[j]);            for (int j=1;j<=n;j++) if (S[j]) lx[j]-=a;            for (int j=1;j<=n;j++) if (T[j]) ly[j]+=a; else sla[j]-=a;        }    }    int ret=0;    for (int i=1;i<=n;i++) if (boy[i]) ret+=w[boy[i]][i];    return ret;}void ToUpperString(string &str){    transform(str.begin(), str.end(), str.begin(), (int (*)(int))toupper);}inline int dcmp(double a,double b){    if (fabs(a-b)<1e-6) return 0;    if (a<b) return -1; return 1;}inline double sqr(double a){ return a*a; }struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) { }    void read() { cin>>x>>y; }    friend Point operator - (Point A,Point B){        return Point(A.x-B.x,A.y-B.y);    }    friend double operator * (Point A,Point B){        return A.x*B.y-B.x*A.y;    }    friend double Dist(Point A,Point B){        return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));    }    friend bool Jud(Point p0,Point p1,Point p2){        double x1,x2;        x1=p1.x; x2=p2.x; if (x1>x2) swap(x1,x2);        if (!(dcmp(x1,p0.x)<=0 && dcmp(p0.x,x2)<=0)) return 0;        x1=p1.y; x2=p2.y; if (x1>x2) swap(x1,x2);        if (!(dcmp(x1,p0.y)<=0 && dcmp(p0.y,x2)<=0)) return 0;        return dcmp((p1-p0)*(p2-p0),0)==0;    }}sx[N],sy[N];map<string,int> Name;double K;int main(){    string name; int iu,iv;    freopen("t.in","r",stdin);    freopen("t.out","w",stdout);    cin>>K>>n;    for (int i=1;i<=n;i++){        sx[i].read(); cin>>name; ToUpperString(name); Name[name]=i;    }    for (int i=1;i<=n;i++){        sy[i].read(); cin>>name; ToUpperString(name); Name[name]=n+i;    }    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) w[i][j]=1;    while (1)    {        cin>>name; if (name==string("End")) break;        ToUpperString(name); iu=Name[name]; cin>>name; ToUpperString(name); iv=Name[name];        if (iu>iv) swap(iu,iv);        iv-=n;        cin>>w[iu][iv];    }    for (int i=1;i<=n;i++)        for (int j=1;j<=n;j++)            for (int k=1;k<=n;k++)            {                if (dcmp(Dist(sx[i],sy[j]),K)>0)                    w[i][j]=-1e8;                else if (k!=i && Jud(sx[k],sx[i],sy[j]))                    w[i][j]=-1e8;                else if (k!=j && Jud(sy[k],sx[i],sy[j]))                    w[i][j]=-1e8;            }    ans=KM();    printf("%d\n",ans);    return 0;}
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