杭电多校联合第一场hdu4606 occupy cities

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4606

题目翻译:在平面上，有n个城市，m个障碍，p个士兵,然后士兵要按照一定的顺序到达城市，士兵在第一次的时候后可以降落在任何一个城市。士兵从城市A 到达城市B 是不能越过障碍，障碍为线段。求士兵走的每路程中每一段的最大值的最小。

分析：前面可以将障碍分成一个个的点，然后判断城市之间的路程与障碍的线段相交，求出每个城市之间的路程，然后就是求图的最小路径覆盖数。二分最大值，按照到达的顺序做成有向边，求最大匹配，然后将城市数目和最大匹配相减与士兵数比较即可（前面的计算几何不会，比赛中没有搞出来。。sad）剩下的都是模板了。。

代码是抄的标程。。。我太水了。。

#include<stdio.h>#include<string.h>#include<algorithm>#include<cmath>#define N 110#define M 210#define eps 1e-8#define CC(x,y) memset(x,y,sizeof(x))#define inf 1e15#define db doubleusing namespace std;struct point{    db x,y;    point() {}    point(db a,db b):x(a),y(b) {}    point operator - (const point p)    {        return point (x-p.x,y-p.y);    }};int dlcmp(double x){    return x<-eps?-1:x>eps;}double sqr(double x){    return x*x;}db dist(point a,point b){    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}db cross(point a,point b){    return a.x*b.y-a.y*b.x;}int n,m,p,seq[N],match[N],map[N][N],link[N],i,j,k,x,y,z;db dis[N+M][N+M];point pt[N+M];int segment_intersect(point s1,point e1,point s2,point e2){    if (max(s1.x,e1.x)>min(s2.x,e2.x)&&max(s2.x,e2.x)>min(s1.x,e1.x)&&            max(s1.y,e1.y)>min(s2.y,e2.y)&&max(s2.y,e2.y)>min(s1.y,e1.y)&&            dlcmp(cross(e1-s1,s2-s1))*dlcmp(cross(e1-s1,e2-s1))<0&&            dlcmp(cross(e2-s2,s1-s2))*dlcmp(cross(e2-s2,e1-s2))<0)        return 1;    return 0;}bool isBlock(point a,point b){    for (int i=1; i<=m; i++)        if (segment_intersect(a,b,pt[i+n],pt[i+n+m]))            return 1;    return 0;}db floyed(){    int i,j,k;    int num=n+m+m;    for (i=1; i<=num; i++)        for (j=1; j<=num; j++)            dis[i][j]=dis[j][i]=(i==j)?0:inf;    for (i=1; i<=num; i++)        for (j=i+1; j<=num; j++)            if (!isBlock(pt[i],pt[j]))                dis[i][j]=dis[j][i]=dist(pt[i],pt[j]);    for (int k=1; k<=num; k++)        for (int i=1; i<=num; i++)            for (int j=1; j<=num; j++)                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);    db res=0;    for (int i=1; i<=n; i++)        for (int j=1; j<=n; j++)            res=max(res,dis[i][j]);    return res;}bool dfs(int x){    int i;    for (i=1; i<=n; i++)        if (match[i]==0&&map[x][i])        {            match[i]=1;            if (link[i]==0||dfs(link[i]))               {                   link[i]=x;                 return 1;               }        }    return 0;}bool check(double x){    CC(map,0);    for (i=1; i<=n; i++)        for (j=1; j<=n; j++)            if (dis[i][j]-x<=eps&&seq[i]<seq[j])                    map[i][j]=1;    CC(link,0);    int ans=0;    for (i=1; i<=n; i++)    {        CC(match,0);        if (dfs(i))ans++;    }    if (n-ans<=p)return 1;    return 0;}int main(){    int tt;    scanf("%d",&tt);    while (tt--)    {        scanf("%d%d%d",&n,&m,&p);        for (int i=1; i<=n; i++)            scanf("%lf%lf",&pt[i].x,&pt[i].y);        for (int i=1; i<=m; i++)            scanf("%lf%lf%lf%lf",&pt[n+i].x,&pt[n+i].y,&pt[n+i+m].x,&pt[n+i+m].y);        for (i=1; i<=n; i++)        {            scanf("%d",&x);            seq[x]=i;        }        db l=0,r=floyed();        while ((r-l)>eps)        {            db mid=(r+l)/2;            if (check(mid))r=mid;            else l=mid;        }        printf("%.2lf\n",l);    }    return 0;}