ZOJ 2710 Two Pipelines

计算几何+贪心

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;



struct City

{

    int id;

    double J1;//距离第一条

    double J2;//距离第二条

    double X,Y;

    double C;

    int be;//属于哪一条

    double xuqiu;



} node[210];

int n,c;

int shu1,shu2;



bool cmpid(const City&a,const City&b) //按照城市编号排序

{

    return a.id<b.id;

}

bool cmpbe(const City&a,const City&b) //按照归属分层

{

    return a.be<b.be;

}

bool cmpC(const City&a,const City&b) //按照差值排序

{

    return a.C<b.C;

}



int main()

{

    int i;

    while(~scanf("%d%d",&n,&c))

    {

        shu1=0,shu2=0;

        double x1,y1,x2,y2;

        double x3,y3,x4,y4;

        double A1,B1,C1;

        double A2,B2,C2;

        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

        scanf("%lf%lf%lf%lf",&x3,&y3,&x4,&y4);

        A1=y2-y1;

        A2=y4-y3;

        B1=x1-x2;

        B2=x3-x4;

        C1=x2*y1-x1*y2;

        C2=x4*y3-x3*y4;

        for(i=0; i<n; i++)

            scanf("%lf%lf%lf",&node[i].X,&node[i].Y,&node[i].xuqiu);

        for(i=0; i<n; i++)

        {

            node[i].J1=fabs(A1*node[i].X+B1*node[i].Y+C1)/sqrt(A1*A1+B1*B1);

            node[i].J2=fabs(A2*node[i].X+B2*node[i].Y+C2)/sqrt(A2*A2+B2*B2);

            if(node[i].J1<=node[i].J2) node[i].be=1,shu1++;

            else node[i].be=2,shu2++;

            node[i].C=fabs(node[i].J1-node[i].J2)*node[i].xuqiu;

            node[i].id=i;

        }

        if(abs(shu1-shu2)<=c)

        {

            for(i=0; i<n; i++)

            {

                if(i<n-1) printf("%d ",node[i].be);

                else printf("%d\n",node[i].be);

            }

        }

        else

        {

            int fenjie=-1;

            sort(node,node+n,cmpbe);

            for(i=0; i<n; i++) if(node[i].be==2) break;

            fenjie=i-1;

            if(shu1>shu2)

            {

                sort(node,node+fenjie+1,cmpC);

                int yidong=0;

                for(i=0; i<=fenjie; i++)

                {

                    node[i].be=2;

                    shu1--;

                    shu2++;

                    if(abs(shu1-shu2)<=c) break;

                }

                sort(node,node+n,cmpid);

                for(i=0; i<n; i++)

                {

                    if(i<n-1) printf("%d ",node[i].be);

                    else printf("%d\n",node[i].be);

                }

            }

            else

            {

                sort(node+fenjie+1,node+n,cmpC);

                int yidong=0;

                for(i=fenjie+1; i<=n-1; i++)

                {

                    node[i].be=1;

                    shu1++;

                    shu2--;

                    if(abs(shu2-shu1)<=c) break;

                }

                sort(node,node+n,cmpid);

                for(i=0; i<n; i++)

                {

                    if(i<n-1) printf("%d ",node[i].be);

                    else printf("%d\n",node[i].be);

                }

            }

        }

    }

    return 0;

}