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;

}

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