nyoj1241 Distribution (河南省第八届acm程序设计大赛)

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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. 

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 2217 14 18 233 156 2830 274 126 78 011 212 255 1019 24

样例输出

-644

来源

第八届河南省程序设计大赛

上传者

hnu_acm

这道题 应该是这里面最简单的一道了把 

就是分割区域 计算坐标在哪个区域里面

#include <stdio.h>
#include <string.h>
struct node
{
	int x,y;
}point[105];
int main()
{
	int n,m,i;
	while(~scanf("%d %d",&n,&m))
	{
		memset(point,0,sizeof(point));
		for(i=0;i<n;i++)
		scanf("%d %d",&point[i].x,&point[i].y);
		for( i=0;i<m;i++)
		{
			int a,b,sum1=0,sum2=0,sum3=0,sum4=0;
			scanf("%d %d",&a,&b);
			for(int i=0;i<n;i++)
			{
				if(point[i].x>a&&point[i].y>b)
				sum1++;
				else if(point[i].x>a&&point[i].y<b)
				sum2++;
				else if(point[i].x<a&&point[i].y>b)
				sum3++;
				else 
				sum4++;
			}
			printf("%d",sum1+sum4-sum2-sum3);
			printf("\n");
		}
	}
	return 0;
}