题目链接：hdu 3642 Get The Treasury

题目大意：三维坐标系，给定若干的长方体，问说有多少位置被覆盖3次以上。

解题思路：扫描线，将第三维分离出来，就是普通的二维扫描线，然后对于每个节点要维护覆盖0，1，2，3以上这4种的覆盖面积。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 4005;

vector<int> pos;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2], s[maxn << 2][5];

inline void pushup(int u) {
    memset(s[u], 0, sizeof(s[u]));

    if (v[u] >= 3)
        s[u][3] = pos[rc[u]+1] - pos[lc[u]];
    else {
        if (lc[u] == rc[u])
            s[u][v[u]] = pos[rc[u]+1] - pos[lc[u]];
        else if (v[u] == 2) {
            s[u][2] = s[lson(u)][0] + s[rson(u)][0];
            for (int i = 1; i <= 3; i++)
                s[u][3] += s[lson(u)][i] + s[rson(u)][i];
        } else if (v[u] == 1) {
            s[u][1] = s[lson(u)][0] + s[rson(u)][0];
            s[u][2] = s[lson(u)][1] + s[rson(u)][1];
            for (int i = 2; i <= 3; i++)
                s[u][3] += s[lson(u)][i] + s[rson(u)][i];
        } else {
            for (int i = 0; i <= 3; i++)
                s[u][i] = s[lson(u)][i] + s[rson(u)][i];
        }
    }
}

inline void maintain(int u, int d) {
    v[u] += d;
    pushup(u);
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    v[u] = 0;

    if (l == r) {
        maintain(u, 0);
        return ;
    }

    int mid = (l + r) / 2;
    build (lson(u), l, mid);
    build (rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int d) {

    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, d);
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, d);
    if (r > mid)
        modify(rson(u), l, r, d);
    pushup(u);
}

struct Seg {
    int x, l, r, d;
    Seg (int x = 0, int l = 0, int r = 0, int d = 0) {
        this->x = x;
        this->l = l;
        this->r = r;
        this->d = d;
    }
};

typedef long long ll;
vector<Seg> g[1005];

inline bool cmp (const Seg& a, const Seg& b) {
    return a.x < b.x;
}

void init () {
    int n, x1, x2, y1, y2, z1, z2;

    scanf("%d", &n);
    for (int i = 0; i <= 1000; i++)
        g[i].clear();

    for (int i = 0; i < n; i++) {
        scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
        for (int i = z1; i < z2; i++) {
            g[i + 500].push_back(Seg(x1, y1, y2, 1));
            g[i + 500].push_back(Seg(x2, y1, y2, -1));
        }
    }
}

inline int find (int k) {
    return lower_bound(pos.begin(), pos.end(), k) - pos.begin();
}

ll solve (int idx) {

    if (g[idx].size() == 0)
        return 0;

    ll ret = 0;
    pos.clear();
    sort(g[idx].begin(), g[idx].end(), cmp);

    for (int i = 0; i < g[idx].size(); i++) {
        pos.push_back(g[idx][i].l);
        pos.push_back(g[idx][i].r);
    }

    sort(pos.begin(), pos.end());
    build(1, 0, pos.size());

    for (int i = 0; i < g[idx].size(); i++) {
        modify(1, find(g[idx][i].l), find(g[idx][i].r) - 1, g[idx][i].d);
        if (i + 1 != g[idx].size())
            ret += 1LL * s[1][3] * (g[idx][i+1].x - g[idx][i].x);
    }
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();

        ll ans = 0;
        for (int i = 0; i <= 1000; i++)
            ans += solve(i);

        printf("Case %d: %I64d\n", kcas, ans);
    }
    return 0;
}

hdu 3642 Get The Treasury(扫描线)