HDU 3642 Get The Treasury(体积并+覆盖三次+线段树+扫描线)

这题就是给出了好多个长方体， 求出那些被覆盖了 3次及以上的体积

我们注意到z的范围很小，而且就给了1000个长方体

那么可以把z坐标离散化

然后在相邻的z坐标空间内，就变成了求面积覆盖

记录三个变量，代表一次覆盖，两次覆盖，三次覆盖及以上

#include <cstdio>#include <cstring>#include <cctype>#include <algorithm>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1#define lch(x) x<<1  #define rch(x) x<<1|1  const int maxn = 2222;int cnt[maxn << 2];//边重复的次数int sum[maxn << 2];//1次及以上int sum2[maxn << 2];//2次及以上int sum3[maxn << 2];int X[maxn],Z[maxn];struct Seg {int h, l, r;int s;//1为入边，-1为出边Seg() {}Seg(int a, int b, int c, int d) : l(a), r(b), h(c), s(d) {}bool operator < (const Seg &cmp) const {return h < cmp.h;}}ss[maxn];struct Spot{int lx, rx, ly, ry, lz, rz;}p[maxn];void PushUp(int rt, int l, int r) {if (cnt[rt] >= 3) sum[rt] = sum2[rt] = 0, sum3[rt] = X[r + 1] - X[l];else if (cnt[rt] == 2){if (l == r) sum2[rt] = X[r + 1] - X[l], sum3[rt] = 0, sum[rt] = 0;else{sum3[rt] = sum[lch(rt)] + sum[rch(rt)] + sum2[lch(rt)] + sum2[rch(rt)] + sum3[lch(rt)] + sum3[rch(rt)];sum2[rt] = X[r + 1] - X[l] - sum3[rt];sum[rt] = 0;}}else if (cnt[rt] == 1){if (l == r) sum[rt] = X[r + 1] - X[l], sum2[rt] = 0, sum3[rt] = 0;else{sum3[rt] = sum2[lch(rt)] + sum2[rch(rt)] + sum3[lch(rt)] + sum3[rch(rt)];sum2[rt] = sum[lch(rt)] + sum[rch(rt)];sum[rt] = X[r + 1] - X[l] - sum2[rt] - sum3[rt];}}else{if (l == r) sum[rt] = sum2[rt] = sum3[rt] = 0;else{sum3[rt] = sum3[lch(rt)] + sum3[rch(rt)];sum2[rt] = sum2[lch(rt)] + sum2[rch(rt)];sum[rt] = sum[lch(rt)] + sum[rch(rt)];}}}void update(int L, int R, int c, int l, int r, int rt) {if (L <= l && r <= R) {cnt[rt] += c;PushUp(rt, l, r);return;}int m = (l + r) >> 1;if (L <= m) update(L, R, c, lson);if (m < R) update(L, R, c, rson);PushUp(rt, l, r);}int main() {int t,cas = 0;scanf("%d", &t);while (t--){int n;scanf("%d", &n);int m = 0;for (int i = 1; i <= n; i++){scanf("%d%d%d%d%d%d", &p[i].lx, &p[i].ly, &p[i].lz, &p[i].rx, &p[i].ry, &p[i].rz);Z[m++] = p[i].lz;Z[m++] = p[i].rz;}//离散化sort(Z, Z + m);int cntz = unique(Z, Z + m) - Z;long long ans = 0;for (int i = 0; i < cntz - 1; i++){memset(cnt, 0, sizeof(cnt));memset(sum, 0, sizeof(sum));memset(sum2, 0, sizeof(sum2));memset(sum3, 0, sizeof(sum3));m = 0;for (int j = 1; j <= n; j++)if (p[j].lz <= Z[i] && p[j].rz > Z[i]){X[m] = p[j].lx;ss[m++] = Seg(p[j].lx, p[j].rx, p[j].ly, 1);X[m] = p[j].rx;ss[m++] = Seg(p[j].lx, p[j].rx, p[j].ry, -1);}sort(X, X + m);sort(ss, ss + m);int cntx = unique(X, X + m) - X;for (int j = 0; j < m - 1; j++){int l = lower_bound(X, X + cntx, ss[j].l) - X;int r = lower_bound(X, X + cntx, ss[j].r) - X - 1;if (l <= r) update(l, r, ss[j].s, 0, cntx - 1, 1);ans += (long long)sum3[1] * (long long)(ss[j + 1].h - ss[j].h) * (long long)(Z[i + 1] - Z[i]);}}printf("Case %d: %lld\n", ++cas, ans);}return 0;}