BZOJ1857: [Scoi2010]传送带(三分套三分)
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Submit: 2005 Solved: 1091
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Description
在一个2维平面上有两条传送带,每一条传送带可以看成是一条线段。两条传送带分别为线段AB和线段CD。lxhgww在AB上的移动速度为P,在CD上的移动速度为Q,在平面上的移动速度R。现在lxhgww想从A点走到D点,他想知道最少需要走多长时间
Input
输入数据第一行是4个整数,表示A和B的坐标,分别为Ax,Ay,Bx,By 第二行是4个整数,表示C和D的坐标,分别为Cx,Cy,Dx,Dy 第三行是3个整数,分别是P,Q,R
Output
输出数据为一行,表示lxhgww从A点走到D点的最短时间,保留到小数点后2位
Sample Input
0 0 0 100
100 0 100 100
2 2 1
100 0 100 100
2 2 1
Sample Output
136.60
HINT
对于100%的数据,1<= Ax,Ay,Bx,By,Cx,Cy,Dx,Dy<=1000
1<=P,Q,R<=10
Source
很显然最优的路线一定是从A走到AB上的一点走到CD上的一点再走到D
然后三分套三分就可以了
#include<cstdio> #include<cmath> #include<algorithm> #define eps 1e-3 using namespace std; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘,c = getchar(); return x * f; } int Ax, Ay, Bx, By, Cx, Cy, Dx, Dy, P, Q, R; double dis(double x, double y, double x2, double y2) { return sqrt((x2 - x) * (x2 -x) + (y2 - y) * (y2 - y)); } double check2(double x,double y,double x2,double y2) { return dis(Ax, Ay, x, y) / P + dis(x, y, x2, y2) / R + dis(x2, y2, Dx, Dy) / Q; } double check(double x, double y) { double lx = Cx, ly = Cy, rx = Dx, ry = Dy, a, b; while(abs(rx - lx) > eps || abs(ry - ly) > eps) { double wx1 = (lx * 2 + rx) / 3, wy1 = (ly * 2 + ry) / 3, wx2 = (lx + rx * 2) / 3, wy2 = (ly + ry * 2) / 3; a = check2(x, y, wx1, wy1); b = check2(x, y, wx2, wy2); if(a > b) lx = wx1, ly = wy1; else rx = wx2, ry = wy2; } return check2(x, y, lx, ly); } int main() { #ifdef WIN32 freopen("a.in","r",stdin); #endif Ax = read(), Ay = read(), Bx = read(), By = read(), Cx = read(), Cy = read(), Dx = read(), Dy = read(), P = read(), Q = read(), R = read(); double lx = Ax, ly = Ay, rx = Bx, ry = By; while(abs(rx - lx) > eps || abs(ry - ly) > eps) { double wx1 = (lx * 2 + rx) / 3, wy1 = (ly * 2 + ry) / 3, wx2 = (lx + rx * 2) / 3, wy2 = (ly + ry * 2) / 3; if(check(wx1, wy1) > check(wx2, wy2)) lx = wx1, ly = wy1; else rx = wx2, ry = wy2; } printf("%.2lf", check(lx, ly)); return 0; }
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