【poj3608】 Bridge Across Islands

http://poj.org/problem?id=3608 (题目链接)

题意：求两凸包间最短距离

Solution 
　　难写难调，旋转卡壳，还真是卡死我了。 
　　先分别选出两凸包最上点和最下点，从这两点开始向逆时针方向旋转卡壳。用叉乘判断是否旋转旋转，具体操作跟求凸包直径差不多。

poj discuss蒯下来的数据制造器：

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;

struct point{double x,y;};
struct polygon
{
    int n;
    point data[10];
};
#define N   16
polygon p[N]={
    {4,{{0,0},{-1,0},{-1,-1},{0,-1}}},
    {4,{{2,0},{2,-1},{3,-1},{3,0}}},
    {4,{{247,208},{247,235},{375,235},{375,208}}},
    {3,{{85 ,101},{116 ,168},{168 ,103}}},
    {3,{{131 ,189},{216 ,148},{196 ,209}}},
    {3,{{180 ,127},{246 ,127},{202 ,144}}},
    {3,{{226 ,201},{297 ,201},{242 ,151}}},
    {3,{{42 ,225},{61 ,261},{100,222}}},
    {3,{{84 ,261},{99 ,246},{102,260}}},
    {3,{{72,309},{157,224},{167,309}}},
    {3,{{170,221},{229,221},{199,245}}},
    {3,{{190,90},{153,47},{225,108}}},
    {3,{{165,230},{168,242},{172,233}}},
    {6,{{143,146},{133,156},{138,170},{150,173},{161,166},{161,152}}},
    {5,{{109,208},{100,236},{111,253},{122,250},{161,206}}},
    {6,{{177,273},{174,309},{202,377},{417,375},{490,270},{268,239}}}
    };
int main()
{
    srand(time(NULL));
    //freopen("aaa.in","w",stdout);
    int i=0,j=0,k,num=0;
    while (i==j) i=rand()%N,j=rand()%N;
    num++;
    //if(num>35||num<=33)continue;
    printf("%d %d\n",p[i].n,p[j].n);
    for(k=0;k<p[i].n;k++)
        printf("%lf %lf\n",p[i].data[k].x,p[i].data[k].y);
    for(k=0;k<p[j].n;k++)
        printf("%lf %lf\n",p[j].data[k].x,p[j].data[k].y);
    printf("0 0\n");
    return 0;
}

代码：

// poj3608
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
#define esp 1e-8
#define inf 2147483640
#define LL long long
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    LL x=0,f=1;char ch=getchar();
    while (ch>‘9‘ || ch<‘0‘) {if (ch==‘-‘) f=-1;ch=getchar();}
    while (ch>=‘0‘ && ch<=‘9‘) {x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}

const int maxn=10010;
struct point {
    double x,y;
    point() {};
    point (double _x,double _y):x(_x),y(_y){}
    friend point operator - (const point &a,const point &b) {
        point x;
        x.x=a.x-b.x;x.y=a.y-b.y;
        return x;
    }
}p1[maxn],p2[maxn],p0;
int sn[maxn],sm[maxn],n,m;

double cross(point p0,point p1,point p2) { //叉乘
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double dis(point a,point b) { //点a与点b之间的距离
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(point a,point b) { //极角排序
    double t=cross(p0,a,b);
    if (t<0) return 0;
    if (t>0) return 1;
    return dis(p0,a)<dis(p0,b);
}
int dcmp(double x) { //double比较
    return fabs(x)<esp ? 0 : (x>0 ? 1 : -1);
}
int Graham(int n,point *p) { //求凸包
    int k=1,top=2;
    for (int i=2;i<=n;i++)
        if (p[i].x==p[k].x ? p[i].y<p[k].y : p[i].x<p[k].x) k=i;
    p0=p[k];p[k]=p[1];p[1]=p0;
    sort(p+2,p+1+n,cmp);
    for (int i=3;i<=n;i++) {
        while (top>1 && cross(p[top-1],p[top],p[i])<=0) top--;
        p[++top]=p[i];
    }
    return top;
}
double pldis(point a,point b,point c) { //点a到线段bc的最短距离
    point s(a-b),t(c-b);
    if (s.x*t.x+s.y*t.y<0) return dis(a,b);
    s=(a-c);t=(b-c);
    if (s.x*t.x+s.y*t.y<0) return dis(a,c);
    return fabs(cross(a,b,c))/dis(b,c);
}
double lldis(point a,point b,point c,point d) { //线段ab与线段cd的最短距离
    return min(min(pldis(a,c,d),pldis(b,c,d)),min(pldis(c,a,b),pldis(d,a,b)));
}
double RC(point *pl,point *pr,int p,int q,int n,int m) { //旋转卡壳
    double tmp,minl=1e90;
    pl[n+1]=pl[1];pr[m+1]=pr[1];
    for (int i=1;i<=n;i++) {
        while ((tmp=cross(pl[p+1],pr[q+1],pl[p])-cross(pl[p+1],pr[q],pl[p]))>esp) q=q%m+1;
        if (tmp<-esp) minl=min(minl,pldis(pr[q],pl[p],pl[p+1]));
        else minl=min(minl,lldis(pl[p],pl[p+1],pr[q],pr[q+1]));
        p=p%n+1;
    }
    return minl;
}
void clocksort() { //逆时针排序
    p0.x=0;p0.y=0;
    for (int i=1;i<=n;i++) {p0.x+=p1[i].x;p0.y+=p1[i].y;}
    p0.x/=n;p0.y/=n;
    sort(p1+1,p1+1+n,cmp);
    p0.x=0;p0.y=0;
    for (int i=1;i<=m;i++) {p0.x+=p2[i].x;p0.y+=p2[i].y;}
    p0.x/=m;p0.y/=m;
    sort(p2+1,p2+1+m,cmp);
}
int main() {
    while (scanf("%d%d",&n,&m)!=EOF && n && m) {
        for (int i=1;i<=n;i++) scanf("%lf%lf",&p1[i].x,&p1[i].y);
        for (int i=1;i<=m;i++) scanf("%lf%lf",&p2[i].x,&p2[i].y);
        n=Graham(n,p1);
        m=Graham(m,p2);
        //clocksort();
        int l=1,r=1;
        for (int i=1;i<=n;i++) if (dcmp(p1[i].y-p1[l].y)<0) l=i; //最下点
        for (int i=1;i<=m;i++) if (dcmp(p2[i].y-p2[r].y)>0) r=i; //最上点
        printf("%.5lf\n",min(RC(p1,p2,l,r,n,m),RC(p2,p1,r,l,m,n)));
    }
    return 0;
}