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;
}