POJ-1556 The Doors 线段相交+最短路

题意：在一个矩形平面内，有若干道墙，现求从左部某一点到右部某一点的最短路径。

解法：有一个事实是线路一定是从门两边的点上通过的，不可能出现从中间穿过的可能。因此我们就枚举两两点之间是否可达，这里就要使用到线段相交的判定。构好图之后就是一个spfa搞定。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

int N;

struct Wall {
    double x, a, b, c, d;
}w[20];

struct Point {
    double x, y;
}e[105];
int idx;

struct Line {
    Point a, b;
    Line(Point x, Point y) {
        a = x, b = y;
    }
    friend bool cross(const Line &, const Line &);
};

double G[105][105];

bool cross(const Line & Line1, const Line & Line2) {
    double Xa1 = Line1.a.x;
    double Ya1 = Line1.a.y;
    double Xa2 = Line1.b.x;
    double Ya2 = Line1.b.y;
    double Xb1 = Line2.a.x;
    double Yb1 = Line2.a.y;
    double Xb2 = Line2.b.x;
    double Yb2 = Line2.b.y;
    if(((Xa2-Xa1)*(Yb1-Ya1)-(Xb1-Xa1)*(Ya2-Ya1))*((Xa2-Xa1)*(Yb2-Ya1)-(Xb2-Xa1)*(Ya2-Ya1))>0)
        return false;
    if(((Xb2-Xb1)*(Ya1-Yb1)-(Xa1-Xb1)*(Yb2-Yb1))*((Xb2-Xb1)*(Ya2-Yb1)-(Xa2-Xb1)*(Yb2-Yb1))>0)
        return false;
    return true;
}


void insert(double x, double y) {
    e[idx].x = x, e[idx].y = y;
    ++idx;
}

bool legal(int x, int y) {
    Line line = Line(e[x], e[y]);
    int l = (x-1)/4, r = (y-1)/4; // 分别计算出这些点属于哪一面墙，再枚举中间的墙
    if (x == 0) l = -1;    // x==0时需特殊处理
    for (int i = l+1; i < r; ++i) { // 计算是否被墙挡住
        if (!cross(line, Line(e[i*4+1], e[i*4+2])) // 如果不从中间墙的某道门穿过的话 
         && !cross(line, Line(e[i*4+3], e[i*4+4]))) {
            return false;
        }
    }
    return true;
}

double dist(const Point & a, const Point & b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

void build() {
    for (int i = 0; i < idx; ++i) {
        for (int j = i + 1; j < idx; ++j) { // 枚举所有的两两组合 
            if (legal(i, j)) {
                G[i][j] = dist(e[i], e[j]);
            }
        }
    }
}

#include <queue>
bool vis[105];
double dis[105];
void spfa() {
    memset(vis, 0, sizeof (vis));
    for (int i = 0; i < idx; ++i) {
        dis[i] = 10000000;    
    }
    dis[0] = 0;
    queue<int>q;
    q.push(0);
    vis[0] = true;
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        vis[v] = false;
        for (int i = v+1; i < idx; ++i) {
            if (G[v][i] != 10000000 && dis[i] > dis[v] + G[v][i]) {
                dis[i] = dis[v] + G[v][i];
                if (!vis[i]) {
                    vis[i] = true;    
                    q.push(i);
                }
            }
        }    
    }
}

int main() {
    while (scanf("%d", &N), N != -1) {
        for (int i = 0; i < 105; ++i) {
            for (int j = 0; j < 105; ++j) {
                G[i][j] = 10000000;    
            }
        }
        idx = 0;
        insert(0.0, 5.0);
        for (int i = 0; i < N; ++i) {
            scanf("%lf %lf %lf %lf %lf", &w[i].x, &w[i].a, &w[i].b, &w[i].c, &w[i].d);
            insert(w[i].x, w[i].a), insert(w[i].x, w[i].b);
            insert(w[i].x, w[i].c), insert(w[i].x, w[i].d);
            // 读取所有的墙，并且添加四个点
        }
        insert(10, 5);
        build();
        spfa();
        printf("%.2f\n", dis[idx-1]);
    }
    return 0;
}