版权声明:欢迎大佬指正! https://blog.csdn.net/sinat_36215255/article/details/82229641
题目网址https://cn.vjudge.net/problem/HihoCoder-1633
比赛的时候只是想到了,将三角形内部的点换成#,然后与三角形严格相交的走法的线段不可行,但是端点相交是可行的。
但是我们忽略了一种情况,就是端点与三角型相交,但是两个端点交于三角形的不同边,这种是错误的,后来我改了之后还想到了一种情况会wa,后来枚举情况,把我hy给搞懵了
看了看人家是怎么解决的(博客),发现人家思路很清晰:
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。。
只要将端点各自向内缩一小块即可。
但是后来发现自己的板子还用错了一个地方,就是判断点在线段上时,端点处应该也要判断一下的,但是我之前没有判断。
好了,终于ac了这个题了。
献上丑陋的代码:
#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
using namespace std;
const int maxn = 25;
struct Point
{
double x,y;
Point(double xx=0,double yy =0)
{
x =xx;
y = yy;
}
// void print()
// {
// cout<<x<<" "<<y<<endl;
// }
};
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);
}
Vector operator/(Vector A,double p)
{
return Vector(A.x/p,A.y/p);
}
bool operator <(const Point &a,const Point &b)
{
return a.x<b.x||(a.x==b.x && a.y<b.y);
}
const double eps = 1e-8;
int dcmp(double x)
{
if(fabs(x)<eps)
return 0;
else
return x<0?-1:1;
}
bool operator == (const Point& a,const Point &b)
{
return dcmp(a.x-b.x)==0 && dcmp(a.y - b.y)==0;
}
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 SegmentProper(Point a1,Point a2,Point b1,Point b2)
{
double c1= Cross(a2-a1,b1-a1),c2 = Cross(a2-a1,b2-a1);
double c3 = Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}
///就是wa下面这里了,还要注意连端点一块判断
bool OnSegment(Point p,Point a1,Point a2)
{
// return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0;
if(dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0 ) return 1;
return a1 == p || a2 == p;
}
Point san[4];
int IsPointIN(Point p)
{
int wn = 0;
int n = 3;
for(int i=0; i<n; i++)
{
if(OnSegment(p,san[i],san[(i+1)%n]))
return -1; //bianejie
int k = dcmp(Cross(san[(i+1)%n]- san[i],p-san[i]));
int d1 = dcmp(san[i].y-p.y);
int d2 = dcmp(san[(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;//niebu
return 0;
}
int n;
char mp[maxn][maxn];
int dx[]= {0,0,1,-1,1,1,-1,-1};
int dy[]= {1,-1,0,0,-1,1,-1,1};
int vis[maxn][maxn];
struct node
{
int x,y;
int t;
node(int xx=0,int yy=0,int tt=0)
{
x =xx;
y = yy;
t =tt;
}
bool operator<(const node & s)const
{
return t>s.t;
}
};
int oj1(int xx,int yy)
{
if(xx>=0&&xx<n&&yy>=0&&yy<n&&!vis[xx][yy] && mp[xx][yy]=='.')
return 1;
return 0;
}
void change(Point &p1,Point &p2,double t)
{
Vector v = p2-p1;
p1 = p1 + v*t;
v = v*(-1.0);
p2 = p2 + v*t;
}
int oj2(int x,int y,int xx,int yy)
{
Point p(x,y);
Point p2(xx,yy);
int ok1=0,ok2=0;
for(int i=0; i<3; i++)
{
if(SegmentProper(p,p2,san[i],san[(i+1)%3]))
return 0;
if(OnSegment(Point(x,y),san[i%3],san[(i+1)%3]))
ok1 = 1;
if(OnSegment(Point(xx,yy),san[i],san[(i+1)%3]))
ok2 = 1;
}
if(ok1&&ok2)
{
if(IsPointIN(Point((double)(xx+x)/2.0,(double)(yy+y)/2.0))==1)
return 0;
}
change(p,p2,1e-3);
if(IsPointIN(p)==1)
return 0;
if(IsPointIN(p2)==1)
return 0;
return 1;
}
int dfs(int a,int b)
{
priority_queue<node> q;
q.push(node(a,b,0));
vis[0][0] = 1;
while(q.size())
{
node a = q.top();
q.pop();
if(a.x==n-1&&a.y==n-1)
{
return a.t;
}
for(int i=0; i<8; i++)
{
int xx = a.x+dx[i];
int yy = a.y+dy[i];
if(!oj1(xx,yy))
continue;
if(!oj2(a.x,a.y,xx,yy))
continue;
vis[xx][yy] = 1;
q.push(node(xx,yy,a.t+1));
}
}
return -1;
}
char xx[maxn][maxn];
int main()
{
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
for(int i=0; i<3; i++)
{
scanf("%lf%lf",&san[i].x,&san[i].y);
}
//cout<<oj2(1,0,1,1)<<endl; //wa的数据
for(int i=n-1; i>=0; i--)
scanf("%s",xx[i]);
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
mp[i][j] = xx[j][i];
}
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
if(IsPointIN(Point(i,j))==1)
{
mp[i][j]='#';
}
}
int ans = dfs(0,0);
printf("%d\n",ans);
}
return 0;
}
/*
3
0 0
1 0.5
2 0
1 0
1 1
*/