NYOJ 1241-Distribution【模拟】

亦凉 2022-08-20 04:12 6阅读 0赞

## 题目1241  [题目信息][Link 1]  [运行结果][Link 2]  [本题排行][Link 3]  [讨论区][Link 4]  Distribution  时间限制： 1000 ms | 内存限制： 65535 KB  难度： 1  描述  One day , Wang and Dong in the Dubai desert expedition, discovered an ancient castle. Fortunately, they found a map of the castle.The map marks the location of treasures.  They agreed to distribute the treasures according to the following rules:        Wang draws a horizontal line on the map and then Dong draws a vertical one so that the map is divided into 4 parts, as show below.  ![20150815154337_96914.jpg][]  Wang will save the treasures in I and III ,while those situated in II and IV will be taken away by Dong. Wang first draw a horizontal line, Dong after the draw a vertical line.  They drew several pairs of lines. For each pair, Wang wants to know the difference between their treasures.  It's guaranteed that all the reasures will lie on neither of the lines drew by them.  输入  the first line contains two integers N and M, where N is the number of treasures on the map and M indicates how many times they are going to draw the lines. The 2nd to (N+1)-th lines Xi, Yi contain the co-ordinates of the treasures and the last M lines consist of the M pairs integers (X, Y) which means that the two splitting lines intersect at point (X, Y).   ( 0 < N, M ≤ 100, 0 ≤ Xi, Yi, X,Y ≤ 1000 )     输出  Output contains M lines , a single line with a integer , the difference described above  样例输入  10 3  29 22  17 14 18 23  3 15  6 28  30 27  4 1  26 7  8 0  11 21  2 25  5 10  19 24  样例输出  \-6  4  4  来源  [第八届河南省程序设计大赛][Link 5]  上传者  [hnu\_acm][hnu_acm]  解题思路：  题目大意就是有一份藏宝图N代表有多少处宝藏，M是划分次数。wang会拿到第一和第三象限的宝藏，Dong会拿到第二和第四象限的宝藏，问wang比dong多拿到多少。  `#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int n,m; struct node { int x,y; }num[10000]; int main() { while(scanf("%d%d",&n,&m)!=EOF) { int i,j; for(i=1;i<=n;i++) { scanf("%d%d",&num[i].x,&num[i].y); } while(m--) { int xx,yy; int u1,u2,u3; u1=u2=u3=0; scanf("%d%d",&xx,&yy); for(i=1;i<=n;i++) { if(xx>num[i].x&&yy>num[i].y) u3++; else if(xx<num[i].x&&yy<num[i].y) u1++; else u2++; } printf("%d\n",u3+u1-u2); } } return 0; }`         ##

[Link 1]: http://acm.nyist.net/JudgeOnline/problem.php?pid=1241
[Link 2]: http://acm.nyist.net/JudgeOnline/status.php?pid=1241
[Link 3]: http://acm.nyist.net/JudgeOnline/problemrank.php?pid=1241
[Link 4]: http://acm.nyist.net/JudgeOnline/talking.php?pid=1241
[20150815154337_96914.jpg]: http://acm.nyist.net/JudgeOnline/admin/kind/attached/20150815154337_96914.jpg
[Link 5]: http://acm.nyist.net/JudgeOnline/search_result.php?source=%E7%AC%AC%E5%85%AB%E5%B1%8A%E6%B2%B3%E5%8D%97%E7%9C%81%E7%A8%8B%E5%BA%8F%E8%AE%BE%E8%AE%A1%E5%A4%A7%E8%B5%9B
[hnu_acm]: http://acm.nyist.net/JudgeOnline/profile.php?userid=hnu_acm