POJ1556计算几何最短路

Posted 2021-10-21 hesorchen

写一道POJ的题就要吐槽一次编译

题目

The Doors

解题思路

对于所有点对，判断能不能建边。

如果有某个墙挡住了这条边，那么不能建边。即判断两线段是否相交，不能算端点。

代码

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 2500;

typedef struct Point
{
    double x, y;
    Point operator-(const Point &temp) const
    {
        Point T;
        T.x = x - temp.x;
        T.y = y - temp.y;
        return T;
    }
} Vector;

double cross(Vector A, Vector B)
{
    return A.x * B.y - A.y * B.x;
}
double dot(Vector A, Vector B)
{
    return A.x * B.x + A.y * B.y;
}
double check(Point a, Point b, Point c, Point d)
{
    Vector V1 = b - a;
    Vector V2 = c - a;
    Vector V3 = d - a;

    Vector V4 = c - d;
    Vector V5 = a - d;
    Vector V6 = b - d;

    if (cross(V1, V3) * cross(V1, V2) < 0 && cross(V4, V5) * cross(V4, V6) < 0)
        return 1;
    return 0;
}
int n;
int ct, cnt;
Point p1[N];
Point p2[N];
Point p[N];

vector<pair<int, double>> vec[N];

double dis[N];
bool vis[N];
void dij()
{
    for (int i = 1; i <= cnt; i++)
        dis[i] = 1e18, vis[i] = 0;
    dis[1] = 0;

    priority_queue<pair<double, int>> q;
    q.push(make_pair(0, 1));

    while (q.size())
    {
        double w = -q.top().first;
        int u = q.top().second;
        q.pop();
        if (vis[u])
            continue;
        vis[u] = 1;
        int m = vec[u].size();
        for (int i = 0; i < m; i++)
        {
            if (!vis[vec[u][i].first])
                if (vec[u][i].second + w < dis[vec[u][i].first])
                {
                    dis[vec[u][i].first] = vec[u][i].second + w;
                    q.push(make_pair(-dis[vec[u][i].first], vec[u][i].first));
                }
        }
    }
    printf("%.2f\\n", dis[cnt]);
}
Point temp;
void solve()
{
    ct = cnt = 0;
    temp.x = 0, temp.y = 5;
    p[++cnt] = temp;
    for (int i = 1; i <= n; i++)
    {
        double x;
        scanf("%lf", &x);
        double y[6];
        for (int i = 1; i <= 4; i++)
        {
            scanf("%lf", &y[i]);
            temp.x = x, temp.y = y[i];
            p[++cnt] = temp;
        }
        temp.x = x, temp.y = 0;
        p1[++ct] = temp;

        temp.x = x, temp.y = y[1];
        p2[ct] = temp;

        temp.x = x, temp.y = y[2];
        p1[++ct] = temp;

        temp.x = x, temp.y = y[3];
        p2[ct] = temp;

        temp.x = x, temp.y = y[4];
        p1[++ct] = temp;

        temp.x = x, temp.y = 10;
        p2[ct] = temp;
    }
    temp.x = 10, temp.y = 5;
    p[++cnt] = temp;
    for (int i = 1; i <= cnt; i++)
        vec[i].clear();
    for (int i = 1; i <= cnt; i++)
    {
        for (int j = i + 1; j <= cnt; j++)
        {
            bool f = 1;
            for (int k = 1; k <= ct; k++)
                if (check(p[i], p[j], p1[k], p2[k])) //是否有线段挡着
                {
                    f = 0;
                    break;
                }
            if (f) //如果没有线段挡着 建边i-j
            {
                vec[i].push_back(make_pair(j, sqrt(dot(p[i] - p[j], p[i] - p[j]))));
                // vec[j].push_back(make_pair(i, sqrt(dot(p[i] - p[j], p[i] - p[j]))));
            }
        }
    }
    dij();//建图完毕，跑最短路
}

int main()
{
    while (scanf("%d", &n) && n != -1)
        solve();

    return 0;
}