HDU 3642 （扫描线、三维体积相交）
题意
在三维空间中给你n个长方体，求空间中被这些长方体覆盖至少3次以上的区域的总体积。
思路
这题没给数据组数T的范围，大致看了一下其他人的都是枚举z来做的，所以我这边也是同样的做法转换成二维的扫描线来做，数组ci表示区间i被覆盖ci次的标记，具体扫描线怎么实现可以看我上篇博客。
1 #define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0) 2 #define bug2(x,y) cout<<#x<<" is "<<x<<" "<<#y<<" is "<<y<<endl 3 #define bug(x) cout<<#x<<" is "<<x<<endl; 4 #include<iostream> 5 #include<algorithm> 6 #include<cstdio> 7 #include<vector> 8 #include<tuple> 9 #define rs o*2+1 10 #define ls o*2 11 using namespace std; 12 typedef long long ll; 13 const int N=2e3+5; 14 int T, kase, n, c[N*4]; 15 ll sum[4][N*4]; 16 vector<tuple<int, int, int, int, int, int>> v, g; 17 vector<int> id; 18 vector<int> idz; 19 void push_up(int o, int l, int r){ 20 if (c[o] >= 3) { 21 for (int i = 1; i <= 3 ;i++) sum[i][o] = id[r] - id[l-1]; 22 } 23 else if (c[o] == 2) { 24 for (int i = 1; i <= 2 ;i++) sum[i][o] = id[r] - id[l-1]; 25 if (l < r) sum[3][o] = sum[1][ls] + sum[1][rs]; 26 else sum[3][o] = 0; 27 } 28 else if (c[o] == 1) { 29 sum[1][o] = id[r] - id[l-1]; 30 if (l < r) { 31 for(int i = 2; i <= 3 ;i++) sum[i][o] = sum[i-1][ls] + sum[i-1][rs]; 32 } 33 else sum[2][o] = sum[3][o] = 0; 34 } 35 else{ 36 if (l < r) { 37 for(int i = 1; i <= 3 ;i++) sum[i][o] = sum[i][ls] + sum[i][rs]; 38 } 39 else sum[1][o] = sum[2][o] = sum[3][o] = 0; 40 } 41 } 42 void up(int o, int l, int r, int ql, int qr, int val){ 43 if (l >= ql && r <= qr) { 44 c[o] += val; 45 push_up(o, l, r); 46 return; 47 } 48 int m = (l+r)/2; 49 if (ql <= m) up(ls, l, m, ql, qr, val); 50 if (qr > m) up(rs, m+1, r, ql, qr, val); 51 push_up(o, l, r); 52 } 53 int gao(int x){ 54 return lower_bound(id.begin(), id.end(), x)-id.begin(); 55 } 56 void init(int n){ 57 for (int i = 1; i <= 8 * n; i++) { 58 c[i] = 0; 59 for (int j = 1; j <= 3; j++) { 60 sum[j][i] = 0; 61 } 62 } 63 } 64 int main(){ 65 IO; 66 cin >> T; 67 while (T--){ 68 cin >> n; 69 id.clear(); 70 v.clear(); 71 int l, r, d, u, h, k; 72 for (int i = 0; i < n; i++){ 73 cin >> l >> d >> k >> r >> u >> h; 74 id.push_back(d); 75 id.push_back(u); 76 idz.push_back(h); 77 idz.push_back(k); 78 v.push_back(make_tuple(l, d, u, 1, k, h)); 79 v.push_back(make_tuple(r, d, u, -1, k, h)); 80 } 81 sort(v.begin(), v.end()); 82 sort(id.begin(), id.end()); 83 id.erase(unique(id.begin(), id.end()), id.end()); 84 sort(idz.begin(), idz.end()); 85 idz.erase(unique(idz.begin(), idz.end()), idz.end()); 86 ll ans = 0; 87 for (int i = 1; i < idz.size(); i++) { 88 int pre = 0; 89 init(n); 90 ll res = 0; 91 g.clear(); 92 for (auto j : v) { 93 int k = get<4>(j), h = get<5>(j); 94 if (idz[i-1] >= k && idz[i] <= h) g.push_back(j); 95 } 96 for (auto j : g) { 97 int pos = get<0>(j), d = get<1>(j), u = get<2>(j), val = get<3>(j); 98 res += 1ll * (pos - pre) * sum[3][1]; 99 pre = pos; 100 up(1, 1, id.size()-1, gao(d)+1, gao(u), val); 101 } 102 ans += res * (idz[i] - idz[i-1]); 103 } 104 printf("Case %d: %I64d\n",++kase,ans); 105 } 106 }
