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
",r);
    }
    return 0;
}