【HDU 3642 求长方体的体积并

来源：互联网发布：有钱人的三观知乎编辑：程序博客网时间：2024/04/28 01:48
思路：很简单就不赘述了，更新父节点的函数注意一下就OK
#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define MAX 3005#define ls rt<<1#define rs ls|1#define m (l+r)>>1int sum1[MAX << 2];int sum2[MAX << 2];int sum3[MAX << 2];int col[MAX << 2];int posx[MAX<<1];int posz[MAX << 1];struct pos{int l, r, h,z1,z2, s;pos(){}pos(int _l, int _r, int _z1,int _z2, int _h, int _s){l = _l;r = _r;h = _h;z1 = _z1;z2 = _z2;s = _s;}bool operator<(pos b){return h < b.h;}}p[MAX << 1],temp[MAX<<1];void uprt(int l,int r,int rt){if (col[rt] == 1){sum3[rt] = sum2[ls] + sum2[rs] + sum3[ls] + sum3[rs];sum2[rt] = sum1[ls] + sum1[rs];sum1[rt] = posx[r + 1] - posx[l] - sum3[rt] - sum2[rt];return;}if (col[rt] == 2){sum3[rt] = sum1[ls]+sum1[rs]+sum2[ls] + sum2[rs] + sum3[ls] + sum3[rs];sum2[rt] = posx[r + 1] - posx[l] - sum3[rt];sum1[rt] = 0;return;}if (col[rt] == 3)//一开始写成==就错了，这个区间覆盖3次以上也是这样的处理，或者不处理，例如，在最下面加上if==0{sum3[rt] = posx[r + 1] - posx[l];sum2[rt] = 0;sum1[rt] = 0;return;}if (col[rt] == 0){sum3[rt] = sum3[ls] + sum3[rs];sum2[rt] = sum2[ls] + sum2[rs];sum1[rt] = sum1[ls] + sum1[rs];return;}}void updata(int L, int R, int c, int l, int r, int rt){if (L <= l&&r <= R){col[rt] += c;uprt(l, r, rt);return;}int mid = m;if (L <= mid)updata(L, R, c, l, mid, ls);if (mid < R)updata(L, R, c, mid + 1, r, rs);uprt(l, r, rt);}int main(){int t;cin >> t;int icase = 1;while (t--){int n;int x1, x2, y1, y2, z1, z2;cin >> n;int cnt = 0;for (int i = 0; i < n; i++){scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);posx[cnt] = x1;p[cnt] = pos(x1, x2, z1,z2, y1, 1);posz[++cnt] = z1;posx[cnt] = x2;p[cnt] = pos(x1, x2, z1,z2, y2, -1);posz[++cnt] = z2;}sort(posx, posx + cnt);sort(posz+1, posz + cnt+1);sort(p, p + cnt);int cntx = unique(posx, posx + cnt) - posx;int cntz = unique(posz+1, posz + cnt+1) - posz-1;long long ans = 0;for (int i = 2; i <=cntz; i++){int cntp = 0;long long disz = posz[i] - posz[i - 1];for (int j = 0; j < cnt;j++)if (p[j].z1<=posz[i - 1]&&p[j].z2>=posz[i])//扫描高，考验扫描线的理解了temp[cntp++] = p[j];memset(sum1, 0, sizeof(sum1));memset(sum2, 0, sizeof(sum2));memset(sum3, 0, sizeof(sum3));memset(col, 0, sizeof(col));for (int j = 0; j < cntp - 1; j++){int curl = lower_bound(posx, posx + cntx, temp[j].l) - posx;int curr = lower_bound(posx, posx + cntx, temp[j].r) - posx-1;updata(curl, curr, temp[j].s, 0, cntx - 1, 1);ans += ((long long)sum3[1]) * (temp[j + 1].h - temp[j].h)*disz;}}printf("Case %d: %lld\n", icase++,ans);}return 0;}
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