nyoj 有趣的问题(最短路径)

经典题目。。

  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 }