POJ 1556 The Doors(Dijkstra+计算几何)

The Doors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3338		Accepted: 1440

Description

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length.

Input

The input data for the illustrated chamber would appear as follows.

2
4 2 7 8 9
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.

Output

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

Source

Mid-Central USA 1996
//Dijkstra+几算几何
//这道题的难点主要是在于如何构建一个图
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double INF = 1 << 30;
const int N = 100;

struct node {
    double x, y;
    int r;
}pnt[N];

struct line {
    node u, v;
}dr[N];

double m[N][N], d[N];   
bool visited[N];        

double dist(node n1, node n2) {
    double x = n1.x - n2.x;
    double y = n1.y - n2.y;
    return sqrt(x * x + y * y);
}

double min(double a, double b) {
    return a < b? a:b;
}

double max(double a, double b) {
    return a > b? a:b;
}

double cross(node n1, node n2, node n0) {
    double x = (n1.x - n0.x) * (n2.y - n0.y);
    double y = (n1.y - n0.y) * (n2.x - n0.x);
    return x - y;
}

bool intersect(node a, node b, node c, node d) {
    bool f1 = (min(a.x, b.x) <= max(c.x, d.x));
    bool f2 = (min(c.x, d.x) <= max(a.x, b.x));
    bool f3 = (min(a.y, b.y) <= max(c.y, d.y));
    bool f4 = (min(c.y, d.y) <= max(a.y, b.y));
    bool f5 = (cross(a, c, d) * cross(c, b, d) >= 0);
    bool f6 = (cross(d, a, b) * cross(a, c, b) >= 0);
        return f1 && f2 && f3 && f4 && f5 && f6;
}

void dijkstra(int v, int n)
{
    int i, j, k;
    if(v < 0 || v > n) return;
    for(i = 0; i <= n; i++) {
        d[i] = m[v][i];
        visited[i] = false;
    }
    d[v] = 0.0; visited[v] = true;
    for(i = 1; i <= n; i++)
    {
        double min = INF;
        k = v;
        for(j = 0; j <= n; j++) {
            if((!visited[j]) && (d[j] < min)) {
                min = d[j];
                k = j;
            }
        }
        visited[k] = true;
        for(j = 0; j <= n; j++) {
            if((!visited[j]) && (m[k][j] < INF)) {
                double newdist = d[k] + m[k][j];
                if(newdist < d[j]) d[j] = newdist;
            }
        }
    }
}
int main()
{
    int i, j, k, n, cnt_1, cnt_2;
    double x, y1, y2, y3, y4;
    while(scanf("%d", &n) && n != -1) {
        cnt_1 = 1; cnt_2 = 1;
        for(i = 1; i <= n; i++)
        {
            scanf("%lf%lf%lf%lf%lf", &x, &y1, &y2, &y3, &y4);
        
                    pnt[cnt_1].x = x; pnt[cnt_1].y = y1; pnt[cnt_1].r = cnt_2; cnt_1++;
                        pnt[cnt_1].x = x; pnt[cnt_1].y = y2; pnt[cnt_1].r = cnt_2; cnt_1++;
                        dr[cnt_2].u = pnt[cnt_1-2]; dr[cnt_2].v = pnt[cnt_1-1]; cnt_2++;

                        pnt[cnt_1].x = x; pnt[cnt_1].y = y3; pnt[cnt_1].r = cnt_2; cnt_1++;
                        pnt[cnt_1].x = x; pnt[cnt_1].y = y4; pnt[cnt_1].r = cnt_2; cnt_1++;
                        dr[cnt_2].u = pnt[cnt_1-2]; dr[cnt_2].v = pnt[cnt_1-1]; cnt_2++;

        }
        pnt[0].x = 0; pnt[0].y = 5; pnt[0].r = 0;
        pnt[cnt_1].x = 10; pnt[cnt_1].y = 5; pnt[cnt_1].r = cnt_2;
        for(i = 0; i <= cnt_1; i++)
            for(j = 0; j <= cnt_1; j++)
                m[i][j] = INF;
        for(i = 0; i <= cnt_1; i++)
        {
            int start = (pnt[i].r+1)/2*2+1;
            for(j = i+1; j <= cnt_1; j++) {
                if(pnt[i].x == pnt[j].x) continue;
                int cnt = 0;
                int end = (pnt[j].r+1)/2*2-1;
                for(k = start; k < end; k++) {
                    if(intersect(pnt[i], pnt[j], dr[k].u, dr[k].v))
                        cnt++;
                }
                if(cnt+1 == (pnt[j].r+1)/2 - (pnt[i].r+1)/2)
                    m[i][j] = m[j][i] = dist(pnt[i], pnt[j]);
            }
        }
        dijkstra(0, cnt_1);
        printf("%.2lf\n", d[cnt_1]);
    }
    return 0;
}