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: Colliding Traffic

For a boat on a small, constrained body of water, other traffic can be a major hazard. The more traffic there is in the same area, the higher the risk of a collision.

Your job is to monitor traffic and help detect likely collisions before they occur. You have sensors to detect the position, direction, and speed of each boat. Assuming the direction and speed remain constant, your task is to determine whether any of the boats will collide. Two boats are considered to collide if they come within a given distance of each other.

Input Specification

The first line of input contains a single integer c, the number of test cases to follow. Each test case starts with a line containing two numbers, n, the number of boats, and r, the collision distance. Two boats are considered to collide if they come within r metres of each other. There will be no more than 1000 boats. Each boat is identified by a line containing four numbers x, y, d, s. The numbers x and y give the current position of the boat as a distance east and north, respectively, from a common origin, and will be between -1000 and 1000, inclusive. The lake is small enough that we can model it as a flat surface. The number d gives the direction in which the boat is heading in degrees clockwise from north (so east is 90 degrees). The number s gives the speed of the boat in metres per second, and will be between 0.001 and 1000. Note that r, x, y, d, and s are not necessarily integers. The input data will be such that the answer will not change if any of the numbers x, y, d and s are changed by 10^-6 or less. If any two boats are within r metres of each other in their initial position, they are already collided, so you should output 0 for that case.

Sample Input

2
2 5
0 0 90 1
10 10 180 1
2 10
0 0 0 0
8 8 270 1

Output Specification

For each test case, output a line containing a single integer, the number of seconds, rounded to the nearest second, before any of the boats come within r metres of each other. If none of the boats ever collide, output the line:

No collision.

Output for Sample Input

6
2

---

Ondřej Lhoták, Richard Peng


題目描述：

給定每艘船的位置以及前向方向和速度，問第一次碰撞的時間，碰撞是看兩點是否在有效範圍內。

題目解法：

窮舉任兩艘可能的碰撞時間取最小即可。
如何決定兩艘船的碰撞時間？
兩條線(船行徑路線)可不必交於一點也可能產生碰撞，平行或重疊也會有追趕的可能。
只考慮兩船的最近距離，則發現必然是呈現凸性(遠-近-遠)或者(遠-無窮遠)。
可利用三分找到最近點的時間，但碰撞可能發生在最近點之前。
再考慮一下，現在已經可以斷出(遠-近) 接著再進行二分找有效範圍的碰撞。
結論：窮舉(三分+二分)

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
struct ship {
    double x, y, d, s;
    double sin, cos;
};
ship S[1005];
int n;
double r;
const double pi = acos(-1);
#define eps 1e-8
double distAB(ship &a, ship &b, double time) {
    static double ax, ay, bx, by;
    ax = a.x + a.s*a.sin*time;
    ay = a.y + a.s*a.cos*time;
    bx = b.x + b.s*b.sin*time;
    by = b.y + b.s*b.cos*time;
    return (ax-bx)*(ax-bx)+(ay-by)*(ay-by);
}
double calcCollide(ship a, ship b, double R) {
    if(distAB(a, b, 0) <= R*R)
        return 0;
    double l, r, mid, midmid;//ternary search
    double md = 1e+30, mmd;
    l = 0, r = 1000;
    while(fabs(l-r) > 1e-2) {
        mid = (l+r)/2;
        midmid = (mid+r)/2;
        md = distAB(a, b, mid);
        mmd = distAB(a, b, midmid);
        if(md <= mmd)    r = midmid;
        else              l = mid;
    }
    if(md > R*R)    return 1e+30;
    // binary search
    r = l, l = 0;
    while(fabs(l-r) > 1e-2) {
        mid = (l+r)/2;
        md = distAB(a, b, mid);
        if(md > R*R)
            l = mid;
        else
            r = mid;
    }
    return l;
}
int main() {
    int testcase;
    int i, j, k;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %lf", &n, &r);
        for(i = 0; i < n; i++) {
            scanf("%lf %lf %lf %lf", &S[i].x, &S[i].y, &S[i].d, &S[i].s);
            S[i].sin = sin(S[i].d/180.0*pi);
            S[i].cos = cos(S[i].d/180.0*pi);
        }
        double ret = 1e+30;
        for(i = 0; i < n; i++) {
            for(j = i+1; j < n; j++) {
                double time = calcCollide(S[i], S[j], r);
                ret = min(ret, time);
            }
        }
        if(ret > 1e+10)
            puts("No collision.");
        else
            printf("%.lf\n", ret);
    }
    return 0;
}