标签:blog os io for ar 问题 div amp
题意:给你N个长方体的左下角和右上角坐标,问你空间中有多少体积是被大于两个不同的立方体覆盖的。x,y~10^6 z~500
考虑到给的z比较小,所以可以直接枚举z,然后跑二维的扫描线就好。
关于处理被不同的线段覆盖三次的问题,可以维护四个信息,cnt,once,twice,more,然后相互推出结果就好。
#include <cstdio> #include <cstring> #include <vector> #include <algorithm> #include <iostream> using namespace std; #define MP make_pair #define PB push_back #define lson rt << 1,l,mid #define rson rt << 1 | 1,mid + 1,r typedef long long LL; typedef vector<int> VI; const int maxn = 4004; struct Seg { int x,l,r,cover; Seg(int x,int l,int r,int cover): x(x),l(l),r(r),cover(cover) {} bool operator < (const Seg &s) const { return x < s.x; } }; VI numz,numy; vector<Seg> s; int n,x1[maxn],y1[maxn],z1[maxn],x2[maxn],y2[maxn],z2[maxn]; int once[maxn << 2],twice[maxn << 2],len[maxn << 2],cnt[maxn << 2]; void getPoint(int &x,int &y,int &z) { scanf("%d%d%d",&x,&y,&z); } int getID(int Val) { return lower_bound(numy.begin(),numy.end(),Val) - numy.begin(); } void pushup(int rt,int l,int r) { int lc = rt << 1,rc = rt << 1 | 1; if(cnt[rt] == 0) { once[rt] = once[lc] + once[rc]; twice[rt] = twice[lc] + twice[rc]; len[rt] = len[lc] + len[rc]; } else if(cnt[rt] == 1) { len[rt] = twice[lc] + twice[rc] + len[lc] + len[rc]; twice[rt] = once[lc] + once[rc]; once[rt] = numy[r + 1] - numy[l] - len[rt] - twice[rt]; } else if(cnt[rt] == 2) { len[rt] = len[lc] + len[rc] + twice[lc] + twice[rc] + once[lc] + once[rc]; once[rt] = 0; twice[rt] = numy[r + 1] - numy[l] - len[rt]; } else { len[rt] = numy[r + 1] - numy[l]; once[rt] = twice[rt] = 0; } } void update(int rt,int l,int r,int ql,int qr,int Val) { if(ql <= l && qr >= r) { cnt[rt] += Val; pushup(rt,l,r); } else { int mid = (l + r) >> 1; if(ql <= mid) update(lson,ql,qr,Val); if(qr > mid) update(rson,ql,qr,Val); pushup(rt,l,r); } } LL solve() { LL ret = 0; int ky = numy.size(),kz = numz.size(); for(int i = 0;i < kz - 1;i++) { s.clear(); for(int j = 0;j < n;j++) if(z1[j] <= numz[i] && z2[j] >= numz[i + 1]) { s.PB(Seg(x1[j],y1[j],y2[j],1)); s.PB(Seg(x2[j],y1[j],y2[j],-1)); } sort(s.begin(),s.end()); int ks = s.size(); LL nw = numz[i + 1] - numz[i]; for(int j = 0;j < ks;j++) { int ql = getID(s[j].l), qr = getID(s[j].r) - 1; update(1,0,ky - 1,ql,qr,s[j].cover); if(j < ks - 1) ret += nw * (s[j + 1].x - s[j].x) * len[1]; } } return ret; } int main() { int T; scanf("%d",&T); for(int kase = 1;kase <= T;kase++) { numy.clear(); numz.clear(); scanf("%d",&n); for(int i = 0;i < n;i++) { getPoint(x1[i],y1[i],z1[i]); getPoint(x2[i],y2[i],z2[i]); numz.PB(z1[i]); numz.PB(z2[i]); numy.PB(y1[i]); numy.PB(y2[i]); } sort(numz.begin(),numz.end()); sort(numy.begin(),numy.end()); numz.erase(unique(numz.begin(),numz.end()),numz.end()); numy.erase(unique(numy.begin(),numy.end()),numy.end()); printf("Case %d: ",kase); cout << solve() << endl; } return 0; }
HDU 3642 线段树+离散化+扫描线,布布扣,bubuko.com
标签:blog os io for ar 问题 div amp
原文地址:http://www.cnblogs.com/rolight/p/3922971.html