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