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HDU 3642 线段树+离散化+扫描线

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,rtypedef 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;}