uvalive 2797 Monster Trap

题意：给定一些线段障碍，判断怪物能不能逃离到无穷远处。

思路：从（0，0）点能否到无穷远处。用BFS搜索。那满足什么样的点符合要求，能加入到图中呢？

遍历每个点，显然一开始已经在某些线段上的点要删去。再判断，两点之间的连线是否与其他线段有交。有则删去。

这道题要注意如果两条线段重合，怎么办？延长每条线段，如果交点在线段内，不算。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
#define up(i,x,y) for(i=x;i<=y;i++)
#define w(a) while(a)
const double inf=0x3f3f3f3f;
const int N = 4010;
const double PI = acos(-1.0);
using namespace std;
const double eps = 1e-12;

double dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Point& A, const Point& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

Vector operator * (const Point& A, double v)
{
    return Vector(A.x*v, A.y*v);
}

Vector operator / (const Point& A, double v)
{
    return Vector(A.x/v, A.y/v);
}

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

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

double Length(const Vector& A)
{
    return sqrt(Dot(A,A));
}

bool operator < (const Point& p1, const Point& p2)
{
    return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}

bool operator == (const Point& p1, const Point& p2)
{
    return p1.x == p2.x && p1.y == p2.y;
}

bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
{
    double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
           c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(const Point& p, const Point& a1, const Point& a2)
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

const int maxv = 200 + 5;
int V;
int G[maxv][maxv], vis[maxv];

bool dfs(int u)
{
    if(u == 1) return true; // 1是终点
    vis[u] = 1;
    for(int v = 0; v < V; v++)
        if(G[u][v] && !vis[v] && dfs(v)) return true;
    return false;
}

const int maxn = 100 + 5;
int n;
Point p1[maxn], p2[maxn];

// 在任何一条线段的中间（在端点不算）
bool OnAnySegment(Point p)
{
    for(int i = 0; i < n; i++)
        if(OnSegment(p, p1[i], p2[i])) return true;
    return false;
}

// 与任何一条线段规范相交
bool IntersectWithAnySegment(Point a, Point b)
{
    for(int i = 0; i < n; i++)
        if(SegmentProperIntersection(a, b, p1[i], p2[i])) return true;
    return false;
}

bool find_path()
{
    // 构图
    vector<Point> vertices;
    vertices.push_back(Point(0, 0)); // 起点
    vertices.push_back(Point(1e5, 1e5)); // 终点
    for(int i = 0; i < n; i++)
    {
        if(!OnAnySegment(p1[i])) vertices.push_back(p1[i]);
        if(!OnAnySegment(p2[i])) vertices.push_back(p2[i]);
    }
    V = vertices.size();
    memset(G, 0, sizeof(G));
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < V; i++)
        for(int j = i+1; j < V; j++)
            if(!IntersectWithAnySegment(vertices[i], vertices[j]))
                G[i][j] = G[j][i] = 1;
    return dfs(0);
}

int main()
{
    while(cin >> n && n)
    {
        for(int i = 0; i < n; i++)
        {
            double x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            Point a = Point(x1, y1);
            Point b = Point(x2, y2);
            Vector v = b - a;
            v = v / Length(v);
            p1[i] = a + v * 1e-6;
            p2[i] = b + v * 1e-6;
        }
        if(find_path()) cout << "no\n";
        else cout << "yes\n";
    }
    return 0;
}