POJ 1556 The Doors --几何，最短路

题意： 给一个正方形，从左边界的中点走到右边界的中点，中间有一些墙，问最短的距离是多少。

解法： 将起点，终点和所有墙的接触到空地的点存下来，然后两两之间如果没有线段（墙）阻隔，就建边，最后跑一个最短路SPFA，即可得出答案。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#define Mod 1000000007
#define eps 1e-8
using namespace std;
#define N 100017

struct Point{
    double x,y;
    Point(double x=0, double y=0):x(x),y(y) {}
    void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
    bool operator < (const Line &L)const { return ang < L.ang; }
};
int dcmp(double x) {
    if(x < -eps) return -1;
    if(x > eps) return 1;
    return 0;
}
template <class T> T sqr(T x) { return x * x;}
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); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
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 Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }

double DisP(Point A,Point B){
    return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
    if(dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) < 0 && dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) < 0) return true;
    return false;
}
//data segment
struct node{
    Point P[2];
}line[206];
Point p[206];
vector<pair<int,double> > G[206];
double dis[206];
int vis[206],tot,Ltot,S,E;
//data ends

void SPFA()
{
    for(int i=1;i<=tot;i++) dis[i] = Mod;
    memset(vis,0,sizeof(vis));
    queue<int> q;
    q.push(S);
    vis[S] = 1, dis[S] = 0;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i=0;i<G[u].size();i++)
        {
            int v = G[u][i].first;
            double w = G[u][i].second;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if(!vis[v]) { q.push(v), vis[v] = 1; }
            }
        }
    }
}

int main()
{
    int n,i,j,k,h;
    double x,a,b,c,d;
    while(scanf("%d",&n)!=EOF && n!=-1)
    {
        tot = 1,Ltot = 0;
        p[1] = Point(0,5);
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf%lf",&x,&a,&b,&c,&d);
            p[++tot] = Point(x,a);
            p[++tot] = Point(x,b);
            p[++tot] = Point(x,c);
            p[++tot] = Point(x,d);
            line[++Ltot].P[0] = Point(x,0), line[Ltot].P[1] = Point(x,a);
            line[++Ltot].P[0] = Point(x,b), line[Ltot].P[1] = Point(x,c);
            line[++Ltot].P[0] = Point(x,d), line[Ltot].P[1] = Point(x,10);
        }
        p[++tot] = Point(10,5);
        Point A,B;
        for(i=0;i<=tot;i++) G[i].clear();
        for(i=1;i<=tot;i++)        //start
        {
            for(j=i+1;j<=tot;j++)  //end
            {
                A = p[i], B = p[j];
                for(k=1;k<=Ltot;k++)
                {
                    if(SegmentIntersection(A,B,line[k].P[0],line[k].P[1]))
                        break;
                }
                if(k == Ltot+1)
                {
                    G[i].push_back(make_pair(j,DisP(A,B)));
                    G[j].push_back(make_pair(i,DisP(A,B)));
                }
            }
        }
        S = 1, E = tot;
        SPFA();
        printf("%.2f\n",dis[E]);
    }
    return 0;
}