题意 : 给你n块矩形,举行只能下降和向左运动,问你他们所占的最小面积是说多少
思路:暴力,我就一直向下向左运动,直到我所有的矩形都运动不了了为止,
注意 : 运动时要检查好边界
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <algorithm>
using namespace std;
struct info{
int x, y, p, q;
}a[1010];
bool cmp1(info a, info b){
return a.y < b.y;
}
bool cmp2(info a, info b){
return a.x < b.x;
}
int main(){
int i, j, k, n, m, x, t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i = 0; i < n; i++)
scanf("%d %d %d %d",&a[i].x, &a[i].y, &a[i].p, &a[i].q);
bool flag = true;
while(flag){
flag = false;
sort(a, a + n, cmp1);
for( int i = 0; i < n; i++ ){
int mx = 0;
for( int j = 0; j < i; j++ ){
if(a[j].q <= a[i].y && !(a[j].x >= a[i].p || a[j].p <= a[i].x))
mx = max(mx, a[j].q);
}
if(mx != a[i].y){
flag = true;
a[i].q -= (a[i].y - mx);
a[i].y = mx;
}
}
sort(a, a + n, cmp2);
for( int i = 0; i < n; i++ ){
int mx = 0;
for( int j = 0; j < i ; j++)
if(a[j].p <= a[i].x && !(a[j].y >= a[i].q || a[i].q <= a[i].y))
mx = max(mx, a[j].p);
if(mx != a[i].x){
flag = true;
a[i].p -=(a[i].x - mx);
a[i].x = mx;
}
}
}
int mx = 0, my = 0;
for( int i = 0; i < n; i++ ){
mx = max(mx, a[i].p);
my = max(my, a[i].q);
}
printf("%d %d\n",mx, my);
}
return 0;
}