Morris' Blog 我在摸索我存在的意、生命的意、涵及值。 有真正神手般的害，套用模式是大部分的情。 那害的我存在的意何，托出神的存在？ 有候不能怪人不我，我已始省思， 什我有人的原因，而是全要靠人。 有候我我不同，那是因我做不到很多人能到的事情。 我不到、玩不起、得不明、理解得不多， 我缺少的文能力，就是一切一切能的判定。 此，我不得能做些人有助的事情。 我知道我自己是不服的性格，但在我不得不。 我有能力，因此我必。 我到底世界作什？充不定性， 而我是法出拔萃，想一般人似乎也回不了了。 越大的挑需要越力的支持，而我的支持是什？ 最近人，但都是失的局，人能接受， 不需要言，不需要通，一切所想都表不出。

49的鼓 6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

#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;
}
 

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