USACO2012 overplanting /// 矩阵切割 递归 oj21547

题目大意：

在农场的任何一个“轴向对齐”的长方形区域（即垂直和水平方向）种植草坪。

现种植了N（1≤ N ≤10）个不同的矩形区域，其中一些甚至可能重叠。

Input

Multiple test cases. For each case:

* Line 1: The integer N.

* Lines 2..1+N: Each line contains four space-separated integers x₁  y₁  x₂  y₂ specifying a rectangular region with upper-left corner (x₁, y₁) and lower-right corner (x₂, y₂).  All coordinates are in the range -10,000...10,000.

Output

For each case, output one line: The total area covered by grass.

Sample Input

2
0 5 4 1
2 4 6 2

Sample Output

20

用矩形切割的方法 看代码就懂

#include <bits/stdc++.h>
using namespace std;
int x1[25],x2[25],m1[25],m2[25],sum;
void cur(int len1,int len2,int hig1,int hig2,int i)
{
    if(len1>=len2||hig1<=hig2) return;
    while(i>=0&&(len1>=x2[i]||hig1<=m2[i]||len2<=x1[i]||hig2>=m1[i]))
        i--;
    if(i<0)
    {
        sum+=(len2-len1)*(hig1-hig2);
        return;
    }
    if(len1<x1[i])
    {
        cur(len1,x1[i],hig1,hig2,i-1);
        len1=x1[i];
    }
    if(len2>x2[i])
    {
        cur(x2[i],len2,hig1,hig2,i-1);
        len2=x2[i];
    }
    if(hig1>m1[i])
    {
        cur(len1,len2,hig1,m1[i],i-1);
        hig1=m1[i];
    }
    if(hig2<m2[i])
    {
        cur(len1,len2,m2[i],hig2,i-1);
        hig2=m2[i];
    }
    return;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        sum=0;
        for(int i=0;i<n;i++)
            scanf("%d%d%d%d",&x1[i],&m1[i],&x2[i],&m2[i]);
        for(int i=0;i<n;i++)
            cur(x1[i],x2[i],m1[i],m2[i],i-1);
        printf("%d
",sum);
    }

    return 0;
}