题目链接：
HDU 2478 Slides
题意：
给出n个矩形的左下角和右上角坐标，求出n-1个矩形相交的最大面积。
分析：
先找到n个矩形左下角中的最右上角和次右上角的x，y坐标(Left_fir.x, Left_fir.y, Left_sec.x, Left_sec.y)，右上角中的最左下角和次左下角的x，y坐标(Right_fir.x, Right_fir.y, Right_sec.x, Right_sec.y)。
其实左下角中的最小右上角x,y和右上角中的最左下角x,y就是n个矩形相交出来的矩形。
然后可以枚举每个矩形，如果第i个矩形的左下角的横坐标是Left_fir.x，那么如果去掉这个矩形，剩下的n个矩形相交的左下角的横坐标是Left_sec.x，否则相交矩形的左下角的横坐标是Left_fir.x，左下角的纵坐标和右上角坐标类似。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 100010;
const int INF = 0x3f3f3f3f;int T, n;
struct Point{int x, y;
}Left[MAX_N], Right[MAX_N], Left_fir, Left_sec, Right_fir, Right_sec;int main()
{scanf("%d", &T);while(T--){scanf("%d", &n);Left_fir.x = Left_fir.y = Left_sec.x = Left_sec.y = -INF;Right_fir.x = Right_fir.y = Right_sec.x = Right_sec.y = INF;for(int i = 0; i < n; ++i){
scanf("%d%d%d%d", &Left[i].x, &Left[i].y, &Right[i].x, &Right[i].y);if(Left[i].x > Left_fir.x){Left_sec.x = Left_fir.x;Left_fir.x = Left[i].x;}else if(Left[i].x > Left_sec.x){Left_sec.x = Left[i].x;}if(Left[i].y > Left_fir.y){Left_sec.y = Left_fir.y;Left_fir.y = Left[i].y;}else if(Left[i].y > Left_sec.y){Left_sec.y = Left[i].y;}if(Right[i].x < Right_fir.x){Right_sec.x = Right_fir.x;Right_fir.x = Right[i].x;}else if(Right[i].x < Right_sec.x){Right_sec.x = Right[i].x;}if(Right[i].y < Right_fir.y){Right_sec.y = Right_fir.y;Right_fir.y = Right[i].y;}else if(Right[i].y < Right_sec.y){Right_sec.y = Right[i].y;}}//printf("left_fir:(%d, %d) left_sec:(%d, %d)\n", Left_fir.x, Left_fir.y, Left_sec.x, Left_sec.y);//printf("right_fir:(%d %d) right_sec:(%d, %d)\n", Right_fir.x, Right_fir.y, Right_sec.x, Right_sec.y);int ans = 0;Point tmp_left, tmp_right;for(int i = 0; i < n; ++i){
if(Left[i].x != Left_fir.x) tmp_left.x = Left_fir.x;else tmp_left.x = Left_sec.x;if(Left[i].y != Left_fir.y) tmp_left.y = Left_fir.y;else tmp_left.y = Left_sec.y;if(Right[i].x != Right_fir.x) tmp_right.x = Right_fir.x;else tmp_right.x = Right_sec.x;if(Right[i].y != Right_fir.y) tmp_right.y = Right_fir.y;else tmp_right.y = Right_sec.y;//printf("tmp_left:(%d, %d) tmp_right:(%d, %d)\n", tmp_left.x, tmp_left.y, tmp_right.x, tmp_right.y);if(tmp_left.x < tmp_right.x && tmp_left.y < tmp_right.y){ans = max(ans, (tmp_right.y - tmp_left.y) * (tmp_right.x - tmp_left.x));}//printf("i = %d ans = %d\n", i, ans);}printf("%d\n", ans);}return 0;
}