hdu 4606 简单计算几何+floyd+最小路径覆盖

思路：将所有的直线的两个端点和城市混在一起，将能直接到达的两个点连线，求一次floyd最短路径。二分枚举bag容量，然后按给的要先后占领的城市由前向后，把能到一步到达的建一条边。然后求一次最小路径覆盖，就是最少需要多少个士兵才能全部占领，跟给出的p值进行比较。

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cmath>

#define Maxn 510

#define Maxm Maxn*Maxn

#define eps 1e-6

#define inf 100000000

using namespace std;

int match[Maxn],vi[Maxn],graphic[Maxn][Maxn],n,m,p,list[Maxn];

double dis[Maxn][Maxn];

struct Point{

    double x,y;

}city[Maxn*10];

struct Edge{

    Point a,b;

}edge[Maxn];

void init()

{

    int i,j;

    memset(vi,0,sizeof(vi));

    memset(match,-1,sizeof(match));

    memset(graphic,0,sizeof(graphic));

    for(i=1;i<=500;i++)

        for(j=1;j<=500;j++)

        dis[i][j]=inf;

}

double multi(Point p0, Point p1, Point p2)//j计算差乘

{   return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}

double Dis(Point &a,Point &b)

{ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}

bool is_cross(Point s1,Point e1,Point s2,Point e2)//判断线段是否相交（非规范相交）

{

   return

          multi(s1,e1,s2)*multi(s1,e1,e2) < -eps &&

          multi(s2,e2,s1)*multi(s2,e2,e1) < -eps;

}

int dfs(int u)

{

    int i,j;

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

    {

        if(!vi[i]&&graphic[u][i])

        {

            vi[i]=1;

            if(match[i]==-1||dfs(match[i]))

            {

                match[i]=u;

                return 1;

            }

        }

    }

    return 0;

}

int maxmatch()

{

    int i,j;

    int ans=0;

    memset(match,-1,sizeof(match));

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

    {

        memset(vi,0,sizeof(vi));

        if(dfs(i))

            ans++;

    }

    return n-ans;

}

int main()

{

    int t,i,j,e,k;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d%d",&n,&m,&p);

        init();

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

            scanf("%lf %lf",&city[i].x,&city[i].y);

        e=n;

        for(i=1;i<=m;i++)

        {

            scanf("%lf%lf%lf%lf",&edge[i].a.x,&edge[i].a.y,&edge[i].b.x,&edge[i].b.y);

            city[++e].x=edge[i].a.x,city[e].y=edge[i].a.y;city[++e].x=edge[i].b.x,city[e].y=edge[i].b.y;

        }

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

            scanf("%d",list+i);

        int flag=0;

        for(i=1;i<e;i++)//将能直线到达的点建边

        {

            for(j=i+1;j<=e;j++)

            {

                flag=0;

                for(k=1;k<=m;k++)

                {

                    if(is_cross(city[i],city[j],edge[k].a,edge[k].b))

                       {

                           flag=1;

                           break;

                       }

                }

                if(!flag)

                dis[i][j]=dis[j][i]=Dis(city[i],city[j]);

            }

        }

        double Max=0;

        for(k=1;k<=e;k++)//求所有点的最短路径

            for(i=1;i<=e;i++)

            for(j=1;j<=e;j++)

            {

            dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);

            }

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

            for(j=1;j<=n;j++)

            if(dis[i][j]!=inf)

            Max=max(Max,dis[i][j]);

        double l,r,mid;

        l=0,r=Max;

        int temp;

        while(r-l>eps)//二分枚举bag容量

        {

            mid=(l+r)/2;

            memset(graphic,0,sizeof(graphic));

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

                for(j=i+1;j<=n;j++)

                {

                    if(dis[list[i]][list[j]]<=mid)

                    {

                        graphic[list[i]][list[j]]=1;

                    }

                }

            temp=maxmatch();

            if(temp>p)

                l=mid;

            else

                r=mid;

        }

        printf("%.2lf\n",r);

    }

    return 0;

}