hdu 3255(扫描线)

题意：有n个人管理农田，每个人管理的农田都是一个矩形，给出每个矩形的左下角坐标和右上角坐标和农田里种的作物，有m种作物，每种作物有自己的生命力和价值，生命力和价值是相等的。每个单位面积的作物只能有一种，生命力最大的作物会生存下来，问价值总值。
题解：如果把每个单位面积的作物的价值当作高，转化为求长方体的体积并。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#define ll long long
using namespace std;
const int N = 60005;
struct Line {
    int lx, rx, h, flag;
    Line(int a, int b, int c, int d): lx(a), rx(b), h(c), flag(d) {}
    bool operator < (const Line& a) const { return h < a.h; }
};
struct Rec {
    int x1, y1, z1, x2, y2, z2;
}rec[N];
int n, m, val[4], flag[N << 2];
ll tree[N << 2];
vector<int> a, b;
vector<Line> line;
map<int, int> mp;

void pushup(int k, int left, int right) {
    if (flag[k])
        tree[k] = a[right] - a[left];
    else if (left + 1 == right)
        tree[k] = 0;
    else
        tree[k] = tree[k * 2] + tree[k * 2 + 1];
}

void modify(int k, int left, int right, int l, int r, int v) {
    if (l <= left && right <= r) {
        flag[k] += v;
        pushup(k, left, right);
        return;
    }
    int mid = (left + right) / 2;
    if (l < mid)
        modify(k * 2, left, mid, l, r, v);
    if (r > mid)
        modify(k * 2 + 1, mid, right, l, r, v);
    pushup(k, left, right);
}

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while (t--) {
        a.clear(), b.clear(), mp.clear();
        memset(tree, 0, sizeof(tree));
        memset(flag, 0, sizeof(flag));
        scanf("%d%d", &n, &m);
        b.push_back(0);
        for (int i = 1; i <= m; i++) {
            scanf("%d", &val[i]);
            b.push_back(val[i]);
        }
        for (int i = 0; i < n; i++) {
            rec[i].z1 = 0;
            scanf("%d%d%d%d%d", &rec[i].x1, &rec[i].y1, &rec[i].x2, &rec[i].y2, &rec[i].z2);
            rec[i].z2 = val[rec[i].z2];
            a.push_back(rec[i].x1);
            a.push_back(rec[i].x2);
        }
        sort(a.begin(), a.end());
        a.erase(unique(a.begin(), a.end()), a.end());
        sort(b.begin(), b.end());
        b.erase(unique(b.begin(), b.end()), b.end());
        int sz = a.size(), sz3 = b.size() - 1;
        for (int i = 0; i < sz; i++)
            mp[a[i]] = i;
        ll res = 0;
        for (int i = 0; i < sz3; i++) {
            line.clear();
            for (int j = 0; j < n; j++) {
                if (rec[j].z1 <= b[i] && rec[j].z2 >= b[i + 1]) {
                    line.push_back(Line(rec[j].x1, rec[j].x2, rec[j].y1, 1));
                    line.push_back(Line(rec[j].x1, rec[j].x2, rec[j].y2, -1));
                }
            }
            sort(line.begin(), line.end());
            int sz2 = line.size();
            for (int j = 0; j < sz2; j++) {
                if (j != 0)
                    res += (b[i + 1] - b[i]) * (line[j].h - line[j - 1].h) * tree[1];
                modify(1, 0, sz - 1, mp[line[j].lx], mp[line[j].rx], line[j].flag);
            }
        }
        printf("Case %d: %lld\n", cas++, res);
    }
    return 0;
}