题解 P1034 【矩形覆盖】

注释非常详细
好几个坐标的运算要多画画图
要注意第64行有一个非常简单的剪枝
然而没有就少了20分
如下贴代码：

代码：

#include<cstdio>
#include<iostream>//其实你也可以考虑用万能但本人不太习惯用，吱吱
using namespace std;
int n, m, ans = 250000 + 2, num;
struct Node{
    int x, y;
};
struct Node node[51];
struct Squre{
    int ax, ay;   //↙左下角(ax, ay) 
    //↘右下角(bx, ay) 
    int bx, by;   //↗右上角(bx, by)
    //↖左上角(ax, by) 
};
struct Squre s[7];
bool is_involve(int i, int j){   //判断点i是否在矩形j内
    if (node[i].x < s[j].ax || node[i].x > s[j].bx) return false;
    if (node[i].y < s[j].ay || node[i].y > s[j].by) return false;
    return true;
}
bool judge(int j){   //判断新矩形j是否影响其他矩形
    for (int i = 1; i <= num; i++){
        if (i == j) continue;
        //方法：只要有某个矩形中有一个点在矩形j内 则有重叠 
        if (s[i].ax >= s[j].ax && s[i].ay >=s[j].ay && s[i].ax <= s[j].bx && s[i].ay <= s[j].by) return false;
        if (s[i].bx >= s[j].ax && s[i].ay >=s[j].ay && s[i].bx <= s[j].bx && s[i].ay <= s[j].by) return false;
        if (s[i].ax >= s[j].ax && s[i].by >=s[j].ay && s[i].ax <= s[j].bx && s[i].by <= s[j].by) return false;
        if (s[i].bx >= s[j].ax && s[i].by >=s[j].ay && s[i].bx <= s[j].bx && s[i].by <= s[j].by) return false;
    }
    return true;
} 
int _s(int i){   //计算第i个矩形的面积 
    return (s[i].by - s[i].ay) * (s[i].bx - s[i].ax);
}
int sum(){   //计算第i个矩形的面积 
    int _sum = 0;
    for (int i = 1; i <= num; i++)
        _sum = _sum + _s(i);
    return _sum;
}
void dfs(int i){   //对于每一个点(node[i].x, node[i].y) 
    //*
    if (i > n){
        ans = min(ans, sum());
//      cout << endl;
//      for (int i = 1; i <= num; i++){
//          cout << s[i].ax << ", " << s[i].ay << "  -  " << s[i].bx << ", " << s[i].by << endl;
//          cout << " + " << _s(i) << endl;
//      }
//      cout << sum() << endl;
        return ;
    }
    if (sum() >= ans) return ;   //Cut
    //1°加入前面的组 
    for (int j = 1; j <= num; j++){
        //1°- 1′可直接加入
        if (is_involve(i, j)){   //如果点i在矩形j内 
            dfs(i + 1);
        }
        else{   //1°- 2′不可直接加入 要扩大原有矩形
            int preax = s[j].ax, preay = s[j].ay, prebx = s[j].bx, preby = s[j].by;
            s[j].ax = min(node[i].x, s[j].ax);
            s[j].ay = min(node[i].y, s[j].ay);
            s[j].bx = max(node[i].x, s[j].bx);
            s[j].by = max(node[i].y, s[j].by);
            if (!judge(j)){   //1°- 2′- 1″扩大原有矩形后 与其他矩形重合 
                s[j].ax = preax; s[j].ay = preay;
                s[j].bx = prebx; s[j].by = preby;   //还原 
                continue;
            }
            //1°- 2′- 2″扩大原有矩形后 不影响其他矩形
            dfs(i + 1); 
            s[j].ax = preax; s[j].ay = preay;
            s[j].bx = prebx; s[j].by = preby;   //还原 
        }
    }
    //2°自成一组 
    if (num < m){
        num++;
        s[num].ax = node[i].x; s[num].bx = node[i].x;
        s[num].ay = node[i].y; s[num].by = node[i].y;
        dfs(i + 1);
        num--;   //还原
    }
    return ;
}

int main(){
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &node[i].x, &node[i].y);
    dfs(1);
    cout << ans;//最后是将答案输出，这个不多说
    return 0;//程序到达了终点，程序：“可以睡觉啦”
}