The Doors 最短路 + 简单几何

　　　　　　　　　　　　　　　　　　　　　　　　The Doors

题目抽象：给你一些点，和一些阻碍连边的线段。问原点到终点最短的几何距离。

分析：预处理见图之后，跑最短路。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 using namespace std;
  6 const int INF = 0X5FFFFFFF;
  7 const int MS = 1005;
  8 const int MS2 = 100;
  9
 10 struct point {
 11     double x, y;
 12    // point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
 13     //长期不更新编译器的oj，可能不支持pioint{}初始化。 可以改为构造函数point()初始化。
 14 }points[MS];
 15 struct edge {
 16     int u, v;
 17     double w;
 18 }edges[MS * MS];
 19
 20 int pSize, eSize;
 21 int n;
 22
 23 double w[MS], wY[MS][4];
 24 double dis[MS];
 25
 26 double D(const point &a, const point &b) {
 27     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 28 }
 29
 30 double C(const point &a, const point &b, const point &c) {
 31     return (c.y - a.y) * (a.x - b.x) - (a.y - b.y) * (c.x - a.x);
 32 }
 33
 34 bool isOk(const point &a,const point &b) {
 35     if (a.x >= b.x)
 36         return false;
 37     int i = 0;
 38     while (i < n && w[i] <= a.x)
 39         i++;
 40     while (i < n && w[i] < b.x ) {
 41         if (C(a, b, point{w[i],0}) * C(a, b, point{w[i],wY[i][0]}) < 0)
 42             return false;
 43         if (C(a, b, point{w[i],wY[i][1]}) * C(a, b, point{w[i], wY[i][2]}) < 0)
 44             return false;
 45         if (C(a, b, point{w[i], wY[i][3]}) * C(a, b, point{w[i], 10.0}) < 0)
 46             return false;
 47         i++;
 48     }
 49     return true;
 50 }
 51
 52 int input() {
 53     scanf("%d", &n);
 54     if (n == -1)
 55         return 0;
 56     pSize = 0;
 57     points[pSize].x = 0;
 58     points[pSize++].y = 5;
 59     for (int i = 0; i < n; i++) {
 60         scanf("%lf", &w[i]);
 61         for (int j = 0;j < 4;j ++) {
 62             points[pSize].x = w[i];
 63             scanf("%lf", &wY[i][j]);
 64             points[pSize++].y = wY[i][j];
 65         }
 66     }
 67     points[pSize].x = 10;
 68     points[pSize++].y = 5;
 69
 70     eSize = 0;
 71     for (int i = 0; i < pSize; i++) {
 72         for (int j = i + 1;j < pSize; j++) {
 73             if (isOk(points[i], points[j])) {
 74                 edges[eSize].u = i;
 75                 edges[eSize].v = j;
 76                 edges[eSize++].w = D(points[i], points[j]);
 77             }
 78         }
 79     }
 80     return 1;
 81 }
 82
 83 void Bellman(int v0) {
 84     for (int i = 0; i < pSize; i++)
 85         dis[i] = INF;     //  fill(dis,dis + pSize, INF )
 86     dis[v0] = 0.0;
 87     int flag = 1;
 88     for (int t = 1; t < pSize && flag; t++) {
 89         flag = 0;
 90         for (int i = 0; i < eSize; i++) {
 91             if (dis[edges[i].u] + edges[i].w < dis[edges[i].v]) {
 92                 dis[edges[i].v] = dis[edges[i].u] + edges[i].w;
 93                 flag = 1;
 94             }
 95         }
 96     }
 97 }
 98
 99 int main() {
100     while (input()) {
101         Bellman(0);
102         printf("%.2lf\n", dis[pSize -1]);
103     }
104     return 0;
105 }