nyoj 有趣的问题(最短路径)

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<cstring>
  5 #include<string>
  6 #include<queue>
  7 #include<algorithm>
  8 #include<map>
  9 #include<iomanip>
 10 #include<climits>
 11 #include<string.h>
 12 #include<cmath>
 13 #include<stdlib.h>
 14 #include<vector>
 15 #include<stack>
 16 #include<set>
 17 #define INF 2000000000
 18 #define MAXN 100
 19 #define maxn 1000010
 20 #define Mod 1000007
 21 #define N 1010
 22 using namespace std;
 23 typedef long long LL;
 24 
 25 struct Point{ double x, y; };
 26 struct Edge{ int u, v; };
 27 int n;
 28 double wX[20];    //每堵墙的坐标
 29 Point p[MAXN];
 30 Edge e[MAXN*MAXN];
 31 int pSize;
 32 double pY[20][4];
 33 double g[MAXN][MAXN];
 34 int eSize;
 35 
 36 //求平面上两点间的距离
 37 double Dis(Point a, Point b)
 38 {
 39     return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
 40 }
 41 
 42 //返回值大于0表示(x3,y3)位于直线上方，小于0表示位于直线下方
 43 double Cross(double x1,double y1,double x2,double y2,double x3,double y3)
 44 {
 45     return (x2 - x1)*(y3 - y1) - (x3 - x1)*(y2 - y1);
 46 }
 47 
 48 bool isOK(Point a, Point b)
 49 {
 50     if (a.x > b.x) return false;
 51     bool flag = true;
 52     int i = 0;
 53     while (wX[i] <= a.x && i < n) i++;
 54     while (wX[i] < b.x && i < n) {
 55         if (Cross(a.x, a.y, b.x, b.y, wX[i], 0)*Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][0]) < 0 || Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][1])*Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][2]) < 0 || Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][3])*Cross(a.x, a.y, b.x, b.y, wX[i], 10) < 0)
 56         {
 57             flag = false;
 58             break;
 59         }
 60         i++;
 61     }
 62     return flag;
 63 }
 64 
 65 //求起始顶点beg到终止顶点end的最短距离
 66 double BellmanFord(int beg, int end)
 67 {
 68     double d[MAXN];
 69     for (int i = 0; i < MAXN; ++i) d[i] = INF;
 70     d[beg] = 0;
 71     bool ex = true;
 72     for (int i = 0; i < pSize && ex; ++i) {
 73         ex = false;
 74         for (int j = 0; j < eSize; ++j) { //判断每条边是否能使顶点v的最短路径距离缩短
 75             if (d[e[j].u] < INF && d[e[j].v] > d[e[j].u] + g[e[j].u][e[j].v])
 76             {
 77                 d[e[j].v] = d[e[j].u] + g[e[j].u][e[j].v];
 78                 ex = true;
 79             }
 80         }
 81     }
 82     return d[end];
 83 }
 84 
 85 void run()
 86 {
 87     p[0].x = 0;
 88     p[0].y = 5;
 89     pSize = 1;
 90     for (int i = 0; i < n; ++i) {
 91         scanf("%lf",&wX[i]);
 92         for (int j = 0; j < 4; ++j) {
 93             p[pSize].x = wX[i];
 94             scanf("%lf", &p[pSize].y);
 95             pY[i][j] = p[pSize].y;
 96             pSize++;
 97         }
 98     }
 99     p[pSize].x = 10;
100     p[pSize].y = 5;
101     pSize++;
102     for (int i = 0; i < pSize; ++i) {
103         for (int j = 0; j < pSize; ++j)
104             g[i][j] = INF;
105     }
106     eSize = 0;
107     for (int i = 0; i < pSize; ++i) {
108         for (int j = i + 1; j < pSize; ++j) {
109             if (isOK(p[i], p[j])) {            //判断第i个点和第j个点是否连线
110                 g[i][j] = Dis(p[i],p[j]);
111                 e[eSize].u = i;
112                 e[eSize].v = j;
113                 eSize++;
114             }
115         }
116     }
117     //求第0个顶点到第pSize-1个顶点之间的最短距离
118     printf("%.2lf\n", BellmanFord(0, pSize - 1));
119 }
120 
121 int main()
122 {
123     while (~scanf("%d", &n)) {
124         if (n == -1) break;
125         run();
126     }
127     return 0;
128 }