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

The famous Korean IT company plans to make a digital map of the Earth with help of wireless sensors which spread out in rough terrains. Each sensor sends a geographical data to . But, due to the inaccuracy of the sensing devices equipped in the sensors, only knows a square region in which each geographical data happens. Thus a geographical data can be any point in a square region. You are asked to solve some geometric problem, known as diameter problem, on these undetermined points in the squares.

A diameter for a set of points in the plane is defined as the maximum (Euclidean) distance among pairs of the points in the set. The diameter is used as a measurement to estimate the geographical size of the set. wants you to compute the largest diameter of the points chosen from the squares. In other words, given a set of squares in the plane, you have to choose exactly one point from each square so that the diameter for the chosen points is maximized. The sides of the squares are parallel to X-axis or Y-axis, and the squares may have different sizes, intersect each other, and share the same corners.

For example, if there are six squares as in the figure below, then the largest diameter is defined as the distance between two corner points of squares S₁ and S₄.

Given a set of n squares in the plane, write a program to compute the largest diameter D of the points when a point is chosen from each square, and to output D², i.e., the squared value of D.

Input 

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. The first line of each test case contains an integer, n, the number of squares, where 2n100, 000. Each line of the next n lines contains three integers, x, y, and w, where (x, y) is the coordinate of the left-lower corner of a square and w is the length of a side of the square; 0x, y10, 000 and 1w10, 000.

Output 

Your program is to write to standard output. Print exactly one line for each test case. The line should contain the integral value D², where D is the largest diameter of the points when a point is chosen from each square.

The following shows sample input and output for two test cases.

Sample Input 

2 
3 
0 0 1 
1 0 2 
0 0 1 
6 
2 1 2 
1 4 2 
3 2 3 
4 4 4 
6 5 1 
5 1 3

Sample Output 

13 
85

---

要找最遠點對, 必須做一次凸包

做法是旋轉卡尺, 不過我想我的做法不是標準的旋轉卡尺, 不過找最遠點對的時候, 仍然是 O(n)

#include <cstdio>
#include <algorithm>
using namespace std;
typedef struct {
    int x, y;
} Pt;
Pt P[500000], H[500000];
bool cmp(Pt a, Pt b) {
    if(a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}
int cross(Pt &o, Pt &a, Pt &b) {
    return (a.x-o.x)*(b.y-o.y) - (a.y-o.y)*(b.x-o.x);
}
void solve(int n) {
    sort(P, P+n, cmp);
    int i, j, m = 0, t;
    for(i = 0; i < n; i++) {
        while(m >= 2 && cross(H[m-2], H[m-1], P[i]) <= 0)
            m--;
        H[m++] = P[i];
    }
    for(i = n-1, t = m+1; i >= 0; i--) {
        while(m >= t && cross(H[m-2], H[m-1], P[i]) <= 0)
            m--;
        H[m++] = P[i];
    }
    m--;
    int ans = 0, tmp, now;
    for(i = 0, j= 1; i < m; i++) {
        while(1) {
            now = (H[i].x-H[j].x)*(H[i].x-H[j].x)+
            (H[i].y-H[j].y)*(H[i].y-H[j].y);
            j++;
            if(j >= m)  j = 0;
            tmp = (H[i].x-H[j].x)*(H[i].x-H[j].x)+
            (H[i].y-H[j].y)*(H[i].y-H[j].y);
            if(tmp < now) {
                j--;
                if(j < 0)   j = m-1;
                break;
            }
        }
        if(now > ans)   ans = now;
    }
    printf("%d\n", ans);
}
int main() {
    int t, x, y, w;
    scanf("%d", &t);
    while(t--) {
        int n, m = 0;
        scanf("%d", &n);
        while(n--) {
            scanf("%d %d %d", &x, &y, &w);
            P[m].x = x, P[m++].y = y;
            P[m].x = x+w, P[m++].y = y;
            P[m].x = x, P[m++].y = y+w;
            P[m].x = x+w, P[m++].y = y+w;
        }
        solve(m);
    }
    return 0;
}