2017ACM-ICPC北京区域现场赛 G题 hihocoder1633

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题意：一个N*N的网格，从左下角走到右上角需要的最短步数。要求不能经过# 而且路径不能经过一个三角形的内部，可以沿边走。

分析：计算几何题里不算难题，但若没想到坑点就一直卡题了。

我采用的判断线段相交方法是规范相交，判断点是否在线段上包括了点在线段端点的情况

假设当前路径是线段L，三角形为ABC

先预处理，若格子在三角形内部，设置为#

case1：L的一个端点在外，一个在内。只要枚举L跟三角形所有边是否规范相交即可。

case2：L的两个端点都在三角形的边上，判断L的中点是否在三角形内部即可

case3：最难想到吧。。一个端点在边上，另一个端点经过了三角形的一个顶点，而且L的中点在三角形外部。。

只要稍微缩短一下L的长度，也就是L的两个端点彼此靠拢很微小的距离，再判断两个端点是否在三角形内部。

还是刘汝佳蓝书的模板用的爽啊~~

AC代码：

#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

const double t = 1e-3;
const double eps = 1e-8;
int dcmp(double x){
    if (fabs(x)<eps) return 0;
    return x<0?-1:1;
}

struct Point{
   double x,y;
   Point(double x=0.0, double y=0.0) : x(x) , y(y) {}
   bool operator ==(const Point & rhs) const {
     return dcmp(x-rhs.x) == 0 && dcmp(y-rhs.y) == 0;
   }
};
Point A,B,C;
vector<Point>poly;

typedef Point Vector;
Vector operator +(Vector A, Vector B) {return Vector(A.x+B.x,A.y+B.y); }
Vector operator -(Vector A, Vector B) {return Vector(A.x-B.x,A.y-B.y); }
Vector operator *(Vector A, double p) {return Vector(A.x*p,A.y*p); }

double Dot(Vector A, Vector B) {return A.x*B.x+A.y*B.y; }
double Cross(Vector A, Vector B) {return A.x*B.y-A.y*B.x; }


bool isPointOnSegment(Point p, Point a1, Point a2) {
    if ( dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p,a2-p))<0 ) return 1;
    return a1 == p || a2 == p;
}

struct Line{
    Point p1,p2;
    Line() {}
    Line(Point p1, Point p2) : p1(p1),p2(p2) {}
    void change(double t) {
       Vector v = p2-p1;
       p1 = p1 + v*t;
       v = v*(-1.0);
       p2 = p2 + v*t;
    }
};
Line AB,AC,BC;

bool LineIntersection(Line a, Line b) {
   double c1 = Cross(a.p2-a.p1,b.p1-a.p1), c2 = Cross(a.p2-a.p1,b.p2-a.p1);
   double c3 = Cross(b.p2-b.p1,a.p1-b.p1), c4 = Cross(b.p2-b.p1,a.p2-b.p1);
   return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

const int maxn = 25;
char s[maxn][maxn];
int n;

int isPointInPolygon(Point p) {
   int wn = 0 ,n = poly.size();
   for (int i=0; i<n; i++) {
      if (isPointOnSegment(p,poly[i],poly[(i+1)%n])) return -1;
      int k = dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));
      int d1 = dcmp(poly[i].y-p.y);
      int d2 = dcmp(poly[(i+1)%n].y-p.y);
      if (k>0 && d1<=0 && d2>0) wn++;
      if (k<0 && d2<=0 && d1>0) wn--;
   }
   if (wn!=0) return 1;
   return 0;
}

void init(){
    double x,y;
    scanf("%lf%lf",&x,&y); A = Point(x,y);
    scanf("%lf%lf",&x,&y); B = Point(x,y);
    scanf("%lf%lf",&x,&y); C = Point(x,y);
    poly.clear();
    AB = Line(A,B); poly.push_back(A);
    BC = Line(B,C); poly.push_back(B);
    AC = Line(A,C); poly.push_back(C);

    char tmp[maxn][maxn];
    for (int i=0; i<n; i++) scanf("%s",tmp[i]);
    for (int i=0; i<n; i++)
       for (int j=0; j<n; j++) s[i][j] = tmp[n-1-j][i];

    for (int i=0; i<n; i++) {
        for (int j=0; j<n; j++) if (s[i][j] == '.') {
             if (isPointInPolygon(Point(i,j)) == 1) s[i][j] = '#';
        }
    }

}

queue<Point>que;
int step[maxn][maxn];
int dx[] = {-1,1,0,0 , -1,-1,1,1};
int dy[] = {0,0,1,-1 , -1,1,-1,1};

int BFS(){
   while (!que.empty()) que.pop();
   memset(step,-1,sizeof(step));
   step[0][0] = 0;

   int x,y;
   int tx,ty;
   Point p,q;
   Point mid,tpoint;
   Vector v;

   p = Point(0.0,0.0); que.push(p);
   while (!que.empty()) {
        p = que.front(); que.pop();
        x = (int)p.x; y = (int)p.y;
        s[x][y] = '#';

        for (int i=0; i<8; i++) {
           tx = x + dx[i];
           ty = y + dy[i];
           if (tx<0 || ty<0 || tx>=n || ty>=n) continue;
           if (s[tx][ty] == '#') continue;
           if (step[tx][ty]!=-1) continue;

           q = Point(1.0*tx,1.0*ty);
           Line L = Line(p,q);
           // case1
           if (LineIntersection(L,AB) || LineIntersection(L,AC) || LineIntersection(L,BC)) continue;

           //casee2
           mid = Point((x+tx)*0.5,(y+ty)*0.5);
           if ( isPointInPolygon(mid) == 1 ) continue;

           // case3
           L.change(t);
           if (isPointInPolygon(L.p1) == 1 || isPointInPolygon(L.p2) == 1) continue;

           step[tx][ty] = step[x][y]+1;
           que.push(q);
        }
   }

   return step[n-1][n-1];
}

int main(){
    while (scanf("%d",&n)==1) {
        init();
        printf("%d\n",BFS());
    }
    return 0;
}