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在一家重要的研究机构工作，小希参与了一项重要的试验：使用激光装置融化巧克力。
该装置有包含 n×m 个单元的一个长方形区域以及一个机械臂组成，每个单元为一个 1×1
的小方块。机械臂上有两束激光垂直射向其表面，任何时候激光束都会射向两个单元的中心。由于两个激光束发射装置装在同一个机械臂上，故移动是同步的，也就是说，移动都在同一方向上。
已知的事实为：
*
*
开始时整个区域被大小为 n×m 的巧克力块覆盖，两束激光均处于区域上方且已激活
*
只有被激光射向单元的巧克力将融化，其他单元内的不受影响
*
机械臂的任何移动必须平行于区域的边缘，每次移动后激光都会同时射向两个单元的中心
*
任何时候激光都只能射向本区域
给出 n 和 m，表示区域的大小，行号从上往下从 1 到 n，列号从左到右为 1 到 m。
给出两束激光最开始时的位置 (x1, y1) 和 (x2, y2），其中 x1、x2 为行号，y1、y2 为列号。
请找出这个区域内有多少单元的巧克力不能被融化。
#include<iostream> #include<stdlib.h> #include<algorithm> using namespace std;
int main() { int t,n,m,x1,y1,x2,y2,x0,y0,xm,ym; cin>>t; for(int i=0;i<t;i++) {
int sum=0; cin>>n>>m>>x1>>y1>>x2>>y2; x0=abs(x1-x2); y0=abs(y1-y2); if(y0==0) {
if(x0<=n/2) {cout<<0<<endl;} if(x0>n/2) { sum=m*n-2*(n-max(x1,x2)+1)*m;
cout<<sum<<endl; } } else if(x0==0) { if(y0<=m/2) {cout<<0<<endl;} if(y0>m/2) {
sum=m*n-2*(m-max(y1,y2)+1)*n; cout<<sum<<endl; } } else { if(x0<=n/2&&y0<=m/2)
{ sum=2*x0*y0; cout<<sum<<endl; } if(x0>n/2&&y0<=m/2) {
sum=n*m-2*(n-max(x1,x2)+min(x1,x2))*(m-max(y1,y2)+min(y1,y2)); cout<<sum<<endl;
} if(x0<=n/2&&y0>m/2) {
sum=n*m-2*(n-max(x1,x2)+min(x1,x2))*(m-max(y1,y2)+min(y1,y2)); cout<<sum<<endl;
} if(x0>n/2&&y0>m/2) {
sum=n*m-2*(n-max(x1,x2)+min(x1,x2))*(m-max(y1,y2)+min(y1,y2)); cout<<sum<<endl;
} } } }
