[UVA] 10488 - Swimming Gopher
Problem E
A Swimming
Gopher
Input: standard input
Output: standard
output
Time Limit: 2 seconds
A gopher
lives in the flatlands of Northern Saskatchewan and travels frequently to see
his cousins. After years and years of drought, there was a period of unusually
heavy rains which resulted in all the ditches on the way of the gopher's travel
being flooded with water. The ditches run in the West-East direction and each
ditch is of a constant width. The gopher still wants to make his travels but being
not a very good swimmer, he wants to minimize the distance that he has to swim
through the ditches and with this requirement satisfied, he also wants to walk
as little as possible on the dry land. Your task is to find the length of the
gopher's route which minimizes his swimming and then his walking.
We are
given a system of orthogonal coordinates, where the x axis is directed
from West to East and the y axis is directed from South to North.
Given are real numbers: D and L, 0 < L, L+L
< D, such that a point (x, y) is on dry land if
and only if there exists an integer n such that
n*D+L <= y <= (n+1)*D-L.
Given is the starting point of the gopher (xs, ys) and its target point (xt, yt). Both these points are on dry land.
Each line of input contains six real numbers: D, L, xs, ys, xt, yt. For each line of input, output one line with the needed information in the format shown in the sample output.
Sample input
100.0 10.0 0.0 50.0 100.0 50.0
100.0 10.0 0.0 50.0 100.0 150.0
100.0 10.0 0.0 -50.0 100.0 50.0
100.0 10.0 0.0 -50.0 100.0 150.0
Output for sample input
The gopher has to swim 0.00 meters and walk 100.00 meters.
The gopher has to swim 20.00 meters and walk 128.06 meters.
The gopher has to swim 20.00 meters and walk 128.06 meters.
The gopher has to swim 40.00 meters and walk 188.68 meters.
Piotr Rudnicki
題目描述：
一隻地鼠要從某一個陸地到另一個陸地，在走水路時的距離越小越好。
保證輸入的兩點一開始都會在陸地上。
題目解法：
計算會跨越多少次水路，經過時一定是垂直穿越，最後將 y 軸距離扣掉，算直角三角形的斜邊即可。
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int main() {
double D, L, xs, ys, xt, yt;
while(scanf("%lf %lf %lf %lf %lf %lf", &D, &L, &xs, &ys, &xt, &yt) == 6) {
int water;
double walk;
water = abs(ceil(ys/D) - ceil(yt/D));
walk = sqrt((xs-xt)*(xs-xt) + (fabs(ys - yt) - water * L * 2)*(fabs(ys - yt) - water * L * 2));
printf("The gopher has to swim %.2lf meters and walk %.2lf meters.\n", water * L * 2, walk);
}
return 0;
}