hdu 3642 Get The Treasury(扫描线)

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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;}

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