基础练习 矩形面积交

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 #define lsn l, mid, rt << 1
 7 #define rsn mid + 1, r, rt << 1 | 1
 8 
 9 //因为只有两个矩形，所以可以不一定要用扫描线线段树的方法
10 //这个简单的方法到时候可以再实现
11 /*
12 typedef double db;
13 struct Scan{
14     double l, r, h; int d;
15     Scan(){}
16     Scan(double l, double r, double h, int d) : l(l), r(r), h(h), d(d){}
17     bool operator < (const Scan & a){
18         return this->h < a.h;  //updata
19     }
20 }scan[100];           //updata
21 double sum[100];
22 double dot[100];
23 int cnt[100];
24 
25 void push_up(int l, int r, int rt){
26     if(cnt[rt]) sum[rt] = dot[r + 1] - dot[l];      //updata
27     else if(l == r) sum[rt] = 0;
28     else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
29 }
30 void update(int L, int R, int v, int l, int r, int rt){
31     if(L <= l && R >= r){
32         cnt[rt] += v;
33         push_up(l, r, rt); //updata
34         return;
35     }
36     int mid = (l + r) >> 1;
37     if(L <= mid) update(L, R, v, lsn);
38     if(R > mid) update(L, R, v, rsn);
39     push_up(l, r, rt);
40 }
41 
42 int main(){
43     int cnt1=0, cnt2 = 0;
44      double ans = 0;
45      for(int i = 1; i < 3; i++){
46         double x1, y1, x2, y2;
47         cin >> x1 >> y1 >> x2 >> y2;
48         ans += (x2 - x1) * (y2 - y1);
49         scan[++cnt1] = Scan(x1, x2, y1, 1);
50         scan[++cnt1] = Scan(x1, x2, y2, -1);
51         dot[++cnt2] = x1; dot[++cnt2] = x2;
52      }
53      sort(scan + 1, scan + 1 + cnt1);
54      sort(dot + 1, dot + 1 + cnt2);
55      int m = 1;
56      for(int i = 2; i <= cnt2; i++){
57         if(dot[i] != dot[i - 1]) dot[++m] = dot[i];
58      }
59      double Ans = 0;
60      //for(int i=1; i<=cnt1; i++) cout << scan[i].h << " ";
61      //cout <<endl;
62      for(int i = 1; i < cnt1; i++){
63         int l = lower_bound(dot + 1, dot + 1 + m, scan[i].l) - dot;
64         int r = lower_bound(dot + 1, dot + 1 + m, scan[i].r) - dot;
65         //cout << l << " " << r << endl;
66         if(l < r) update(l, r - 1, scan[i].d, 1, m, 1);
67         //updata
68         Ans += sum[1] * (scan[i + 1].h - scan[i].h);
69         //cout << Ans << " " << sum[1] << " " << i <<  " " << scan[i+1].h << " " << scan[i].h << " " << scan[i + 1].h - scan[i].h << endl;
70      }
71     //cout << ans << " " << Ans << endl;
72     printf("%.2f\n", ans - Ans);
73     return 0;
74 }
75 */
76 //这里有个坑。。。输入并不一定是先输入矩形的左下角再输入矩形的右下角，所以在判断是否有交之前我们要先sort一下，将左下角的坐标放在前面的位置
77 //不得不服别人的智商。。。很巧妙啊
78 double x[4], y[4];
79 int main(){
80     for(int  i = 0; i < 4; i += 2){
81         cin >> x[i] >> y[i] >> x[i+1] >> y[i+ 1];
82     }
83     sort(x, x+2);
84     sort(x+2, x+4);
85     sort(y, y+2);
86     sort(y+2, y+4);
87     if(x[0] >= x[3] || x[1] <= x[2] || y[0] >= y[3] || y[1] <= y[2]){
88         printf("0.00\n");
89     }else{
90         sort(x, x+4);
91         sort(y, y+4);
92         printf("%.2lf\n", (x[2] - x[1]) * (y[2] - y[1]) );
93     }
94     return 0;
95 }