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 A: Brownie Points I

Stan and Ollie play the game of Odd Brownie Points. Some browniepoints are located in the plane, at integer coordinates. Stan playsfirst and places a vertical line in the plane. The line must gothrough a brownie point and may cross many (with the samex-coordinate). Then Ollie places a horizontal line that must crossa brownie point already crossed by the vertical line.

Those lines divide the plane into four quadrants. The quadrantcontaining points with arbitrarily large positive coordinates is thetop-right quadrant.

The players score according to the number of brownie points in thequadrants. If a brownie point is crossed by a line, it doesn'tcount. Stan gets a point for each (uncrossed) brownie point in thetop-right and bottom-left quadrants. Ollie gets a point for each(uncrossed) brownie point in the top-left and bottom-right quadrants.

Your task is to compute the scores of Stan and Ollie given the pointthrough which they draw their lines.

Input contains a number of test cases. The data of each test caseappear on a sequence of input lines. The first line of each test casecontains a positive odd integer1 < n < 200000 which is the numberof brownie points. Each of the following n lines containstwo integers, the horizontal (x) and vertical (y)coordinates of a brownie point. No two brownie points occupy the sameplace. The input ends with a line containing 0 (instead of then of a test).

For each test case of input, print a line with two numbers separatedby a single space. The first number is Stan's score, the second is the score of Ollie when their lines cross the point whose coordinatesare given at the center of the input sequence of points for this case.

Sample input

113 23 33 43 62 -21 -30 0-3 -3-3 -2-3 -43 -70

Output for sample input

6 3

---

P. Rudnicki

題目解法：
這題乍看起來有點近似於博弈問題，一個人先決定垂直線，接著換另一個人決定水平線。
接著畫分成四個象限，一三象限的點個數給第一個人，其餘給第二個人，點數即分數。

但是那是事實上不用，題目已經給定好輸入序列中的中間點做為畫分對象。


//博弈的話，要正常的窮舉，即是窮舉所有點的分差最低的那個。


#include <stdio.h>
#include <algorithm>
using namespace std;
struct Pt {
    int x, y;
};
Pt D[200000];
int main() {
    int n, i;
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++)
            scanf("%d %d", &D[i].x, &D[i].y);
        int mx, my;
        int stan = 0, olli = 0;
        mx = D[n/2].x, my = D[n/2].y;
        for(i = 0; i < n; i++) {
            if(D[i].x > mx) {
                if(D[i].y > my)
                    stan++;
                if(D[i].y < my)
                    olli++;
            }
            if(D[i].x < mx) {
                if(D[i].y > my)
                    olli++;
                if(D[i].y < my)
                    stan++;
            }
        }
        printf("%d %d\n", stan, olli);
    }
    return 0;
}