Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem C: Snow Clearing

As the days become shorter and the nights become longer we turn our thoughts to snow clearing. Due to budget cuts, the Big Noisy City has exactly one snow plow. The plow can clear exactly one lane of a road in a single pass. Whenever there is snow on the ground, the plow departs from its hangar and tours the city, plowing as it goes. What is the minimum time that the plow needs to clear every lane of every road?

The first line of input is the number of test cases, followed by a blank line. Then the input contains two integers: the x,y coordinates of the hangar (in metres). Up to 100 lines follow. Each gives the coordinates (in metres) of the beginning and end of a street. All roads are perfectly straight, with one lane in each direction. The plow can turn any direction (including a U-turn) at any intersection, and can turn around at the end of any street. The plow travels at 20 km/h if it is plowing, and 50 km/h if the lane it is driving on has already been plowed. It is possible to reach all streets from the hangar. There is a blank line between each consecutive test case.

Your output should be the time, in hours and minutes, required to clear the streets and return to the hangar. Round to the nearest minute  for each data set. Print a blank line between 2 consecutive data sets.

Sample Input

1

0 0
0 0 10000 10000
5000 -10000 5000 10000
5000 10000 10000 10000

Output for Sample Input

3:55

---

題目已經偷偷地透露要經過所有邊，且回到一開始出發的地方，求最少花費時間。

但題目又偷偷表示，每一條道路都有兩個方向。

在有向圖中，歐拉迴路符合所有點的入度等於出度，因此直接計算所有道路的長度即可。


#include <stdio.h>
#include <math.h>
int main() {
    int testcase;
    char s[1024];
    scanf("%d", &testcase);
    while(getchar() != '\n');
    while(getchar() != '\n');
    while(testcase--) {
        gets(s);
        double x1, y1, x2, y2;
        double dist = 0;
        while(gets(s) && s[0] != '\0') {
            sscanf(s, "%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
            dist += hypot(x1 - x2, y1 - y2);
        }
        dist = dist / 1000; // km
        double hh = dist * 2 / 20; // * 2 (with one lane in each direction)
        int mm = (int)round(hh * 60);
        printf("%d:%02d\n", mm/60, mm%60);
        if(testcase)
            puts("");
    }
    return 0;
}