NYOJ_5743Distribution（第八届河南省程序设计大赛）

题目描述
One day , Wang and Dong in the Dubai desertexpedition, discovered an ancient castle. Fortunately, they found a map of thecastle.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 betaken away by Dong. Wangfirst draw a horizontal line, Dong after the draw a vertical line.
They drewseveral pairs of lines. For each pair, Wangwants to know the difference between their treasures.

It's guaranteed that all the reasures will lie on neither of the linesdrew 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

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;struct node{    int x,y;};int main(){    int n,m,i,j,a,b,s4,s2,x,y;    node mm[1100];    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%d%d",&mm[i].x,&mm[i].y);        }        for(i=0;i<m;i++)        {            s4=s2=0;            scanf("%d%d",&a,&b);            for(j=0;j<n;j++)            {                if((mm[j].x>a&&mm[j].y>b)||(mm[j].x<a&&mm[j].y<b))                    s4++;                else                    s2++;            }            printf("%d\n",s4-s2);        }    }    return 0;}