poj 2966 A safe way

WA在那放了一年了，现在再做感觉思路清晰多了。

题意，求多边形外两点不经过多边形内部的最短路，可擦边过。

先枚举判断任意两点间线段是否会经过多边形内部，再用Dijkstra求单源最短路。

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;const double EP=1e-8;const double inf=1e100;inline double max(double x, double y){ return x>y?x:y;}inline double min(double x, double y){ return x<y?x:y;}int n, cnt, vis[110];double a[110][110], d[110];struct Point{    double x, y;}p[110], tp[110], t1, t2;struct Line{    Point a, b;};bool cmpx(Point p1, Point p2){    return p1.x<p2.x;}bool cmpy(Point p1, Point p2){    return p1.y<p2.y;}double x_mult(Point sp, Point ep, Point op){    return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);}double dist(Point p1, Point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}Point intersection(Point u1,Point u2,Point v1,Point v2){      Point res=u1;      double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/               ((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));     res.x+=(u2.x-u1.x)*t;     res.y+=(u2.y-u1.y)*t;     return res;}bool Point_In_Segment(Point a, Point b, Point t){       //判断点在线段上    return (t.x<=max(a.x, b.x)&&            t.x>=min(a.x, b.x)&&            t.y<=max(a.y, b.y)&&            t.y>=min(a.y, b.y)&&            fabs(x_mult(t, a, b))<EP);}bool Point_In_Polygon(Point pt){                        //判断点在多边形内（包括边界）    int i, num=0;    for(i=0; i<n; i++){        Point p1=p[i], p2=p[(i+1)%n];        if(Point_In_Segment(p1, p2, pt))return true;        if(p1.y==p2.y||pt.y<min(p1.y, p2.y)||pt.y>=max(p1.y, p2.y))        continue;        double t=(pt.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x;        if(t>pt.x)num++;    }    return num&1;}bool Linelenth_In_Polygon(Line l1){        //求线段在多边形内长度（不包括边界）    int i, j;    double ans=0;    Point mid;    cnt=0;    for(i=0; i<n; i++){        if((x_mult(p[i], l1.a, l1.b)*x_mult(p[(i+1)%n], l1.a, l1.b)<=EP)&&        (x_mult(l1.a, p[i], p[(i+1)%n])*x_mult(l1.b, p[i], p[(i+1)%n])<=EP)){            if(fabs(x_mult(p[i], l1.a, l1.b))<EP&&fabs(x_mult(p[(i+1)%n], l1.a, l1.b))<EP)  //重合时无交点处理continue;            tp[cnt++]=intersection(p[i], p[(i+1)%n], l1.a, l1.b);        }    }if(fabs(l1.a.x-l1.b.x)>EP)      //选择按x还是y排序，防止平行坐标轴时精度问题sort(tp, tp+cnt, cmpx);else sort(tp, tp+cnt, cmpy);    for(i=0; i<cnt-1; i++){        mid.x=(tp[i].x+tp[i+1].x)/2;        mid.y=(tp[i].y+tp[i+1].y)/2;        int flag=0;        for(j=0; j<n; j++){            if(Point_In_Segment(p[j], p[(j+1)%n], mid))     //去掉与边重合的部分            flag=1;        }        if(flag)continue;        if(Point_In_Polygon(mid))        ans+=dist(tp[i], tp[i+1]);        if(ans>EP)return true;    }    return false;}double dijkstra(int st){    memset(vis, 0, sizeof(vis));    int i, j;    for(i=0; i<=n+1; i++){        if(a[st][i]<0)d[i]=inf;        else d[i]=a[st][i];    }    d[st]=0;    vis[st]=1;    for(i=0; i<=n; i++){        double tmp=inf;        int flag;        for(j=0; j<=n+1; j++){            if(!vis[j]&&d[j]<tmp){                tmp=d[j];                flag=j;            }        }        if(tmp==inf)break;        vis[flag]=1;        for(j=0; j<=n+1; j++){            if(a[j][flag]>=0&&!vis[j]&&d[j]>d[flag]+a[j][flag])            d[j]=d[flag]+a[j][flag];        }    }    //for(i=0; i<=n+1; i++)printf("%lf\n", d[i]);    return d[n+1];}void read(){    int i, j;    Line l1;    scanf("%d", &n);    for(i=0; i<n; i++)        scanf("%lf%lf", &p[i].x, &p[i].y);    p[n]=t1;p[n+1]=t2;    for(i=0; i<=n; i++)    for(j=i+1; j<=n+1; j++){        l1.a=p[i];l1.b=p[j];        if(Linelenth_In_Polygon(l1))a[i][j]=a[j][i]=-1;        else a[i][j]=a[j][i]=dist(p[i], p[j]);    }    printf("%.4f\n", dijkstra(n));}int main(){    //freopen("1.txt", "r", stdin);    while(scanf("%lf%lf%lf%lf", &t1.x, &t1.y, &t2.x, &t2.y)!=EOF){        read();    }    return 0;}