题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1255
题意很清楚，就是让你求矩阵之间叠加层数大于1的矩形块的面积和。
因为n只有1000，所以我离散化一下，数据大小就缩小了，那么之后只需要线段树单点更新就好了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <map> 5 #include <algorithm> 6 using namespace std; 7 const int MAXN = 4e3 + 5; 8 struct data { 9 int x1 , x2 , flag; 10 double y , xx1 , xx2; 11 bool operator <(const data& cmp) const { 12 return y < cmp.y; 13 } 14 }a[MAXN]; 15 struct segtree { 16 int l , r , add; 17 double val; 18 }T[MAXN << 2]; 19 double x[MAXN]; 20 map <double , int> mp; 21 22 void build(int p , int l , int r) { 23 int mid = (l + r) >> 1; 24 T[p].val = T[p].add = 0 , T[p].l = l , T[p].r = r; 25 if(r - l == 1) { 26 return ; 27 } 28 build(p << 1 , l , mid); 29 build((p << 1)|1 , mid , r); 30 } 31 32 void updata(int p , int l , int r , int val) { 33 int mid = (T[p].l + T[p].r) >> 1; 34 if(T[p].r - T[p].l == 1) { 35 T[p].add += val; 36 if(T[p].add > 1) 37 T[p].val = x[T[p].r] - x[T[p].l]; 38 else 39 T[p].val = 0; 40 return ; 41 } 42 if(r <= mid) { 43 updata(p << 1 , l , r , val); 44 } 45 else if(l >= mid) { 46 updata((p << 1)|1 , l , r , val); 47 } 48 else { 49 updata(p << 1 , l , mid , val); 50 updata((p << 1)|1 , mid , r , val); 51 } 52 T[p].val = T[p << 1].val + T[(p << 1)|1].val; 53 } 54 55 int main() 56 { 57 int t = 1 , n; 58 double x1 , x2 , y1 , y2; 59 scanf("%d" , &t); 60 while(t--) { 61 mp.clear(); 62 scanf("%d" , &n); 63 int cnt = 0; 64 for(int i = 0 ; i < n ; i++) { 65 scanf("%lf %lf %lf %lf" , &x1 , &y1 , &x2 , &y2); 66 if(x1 > x2) 67 swap(x1 , x2); 68 if(y1 > y2) 69 swap(y1 , y2); 70 int ls = i * 2 , rs = i * 2 + 1; 71 a[ls].xx1 = x1 , a[ls].xx2 = x2 , a[ls].y = y1 , a[ls].flag = 1; 72 a[rs].xx1 = x1 , a[rs].xx2 = x2 , a[rs].y = y2 , a[rs].flag = -1; 73 if(!mp[x1]) { 74 mp[x1] = 1; 75 x[++cnt] = x1; 76 } 77 if(!mp[x2]) { 78 mp[x2] = 1; 79 x[++cnt] = x2; 80 } 81 } 82 sort(a , a + n * 2); 83 sort(x + 1 , x + cnt + 1); 84 for(int i = 0 ; i < n * 2 ; i++) { 85 a[i].x1 = lower_bound(x + 1 , x + cnt + 1 , a[i].xx1) - x; 86 a[i].x2 = lower_bound(x + 1 , x + cnt + 1 , a[i].xx2) - x; 87 } 88 double res = 0; 89 build(1 , 1 , cnt); 90 updata(1 , a[0].x1 , a[0].x2 , a[0].flag); 91 for(int i = 1 ; i < 2 * n ; i++) { 92 res += (a[i].y - a[i - 1].y) * T[1].val; 93 updata(1 , a[i].x1 , a[i].x2 , a[i].flag); 94 } 95 printf("%.2f " , res); 96 } 97 }