Get The Treasury HDU - 3642（体积扫描线）

给出n个立方体，要你求这些立方体至少被覆盖三次的部分。

先把这个立方体的信息存在来，发现Z的范围不大，z范围是是[-500,500]，所以我们可以先离散化，然后枚举Z，

然后对于每一段Z的区域内，在当前的区域内对xoy轴使用一次扫描线，找到当前这个区域内被覆盖三次的体积，然后每次枚举Z，每次相加，就是最后的答案。

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define first fi
#define second se
#define lowbit(x) (x & (-x))

typedef unsigned long long int ull;
typedef long long int ll;
const double pi = 4.0*atan(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 2050;
const int maxm = 305;
using namespace std;

int n, m, tol, T;
struct Node {
    int l, r, h, f;
    bool operator < (Node a) const {
        return h < a.h;
    }
};
Node node[maxn];
struct Edge {
    int x1, y1, z1;
    int x2, y2, z2;
};
Edge edge[maxn];
int Z[maxn];
int a[maxn];
int cnt[maxn << 2];
int sum1[maxn << 2];
int sum2[maxn << 2];
int sum3[maxn << 2];

void init() {
    m = 1;
    memset(cnt, 0, sizeof cnt);
    memset(sum1, 0, sizeof sum1);
    memset(sum2, 0, sizeof sum2);
    memset(sum3, 0, sizeof sum3);
    memset(node, 0, sizeof node);
}

void pushup(int left, int right, int root) {
    if(cnt[root] >= 3) {
        sum3[root] = sum2[root] = sum1[root] = a[right+1] - a[left];
    } else if(cnt[root] == 2) {
        sum2[root] = sum1[root] = a[right+1] - a[left];
        sum3[root] = sum1[root << 1] + sum1[root << 1 | 1];
    } else if(cnt[root] == 1) {
        sum1[root] = a[right+1] - a[left];
        sum3[root] = sum2[root << 1] + sum2[root << 1 | 1];
        sum2[root] = sum1[root << 1] + sum1[root << 1 | 1];
    } else {
        sum3[root] = sum3[root << 1] + sum3[root << 1 | 1];
        sum2[root] = sum2[root << 1] + sum2[root << 1 | 1];
        sum1[root] = sum1[root << 1] + sum1[root << 1 | 1];
    }
}

void update(int left, int right, int prel, int prer, int val, int root) {
    if(prel <= left && right <= prer) {
        cnt[root] += val;
        pushup(left, right, root);
        return ;
    }
    int mid = (left + right) >> 1;
    if(prel <= mid)    update(left, mid, prel, prer, val, root << 1);
    if(prer > mid)    update(mid+1, right, prel, prer, val, root << 1 | 1);
    pushup(left, right, root);
}

int main() {
    int T;
    scanf("%d", &T);
    int cas = 1;
    while(T--) {
        memset(edge, 0, sizeof edge);
        memset(Z, 0, sizeof Z);
        scanf("%d", &n);
        for(int i=1; i<=n; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            edge[i].x1 = x;
            edge[i].y1 = y;
            edge[i].z1 = z;
            Z[2*i-1] = z;
            scanf("%d%d%d", &x, &y, &z);
            edge[i].x2 = x;
            edge[i].y2 = y;
            edge[i].z2 = z;
            Z[2*i] = z;
        }
        sort(Z+1, Z+1+2*n);
        int zz = unique(Z+1, Z+1+2*n) - (Z+1);
        ll ans = 0;
        for(int i=1; i<zz; i++) {
            init();
            for(int j=1; j<=n; j++) {
                if(edge[j].z1 <= Z[i] && Z[i+1] <= edge[j].z2) {
                    node[m].l = edge[j].x1;
                    node[m].r = edge[j].x2;
                    node[m].h = edge[j].y1;
                    node[m].f = 1;
                    a[m++] = edge[j].x1;
                    node[m].l = edge[j].x1;
                    node[m].r = edge[j].x2;
                    node[m].h = edge[j].y2;
                    node[m].f = -1;
                    a[m++] = edge[j].x2;
                }
            }
            m--;
            sort(node+1, node+1+m);
            sort(a+1, a+1+m);
            int nn = unique(a+1, a+1+m) - (a+1);
            for(int j=1; j<m; j++) {
                int l = lower_bound(a+1, a+1+nn, node[j].l) - a;
                int r = lower_bound(a+1, a+1+nn, node[j].r) - a;
                update(1, m, l, r-1, node[j].f, 1);
                ans += 1ll * sum3[1] * (node[j+1].h - node[j].h) * (Z[i+1] - Z[i]);
            }
        }
        printf("Case %d: %lld\n", cas++, ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/H-Riven/p/9360502.html