HDU 3255 farming (scanning line)

Link: HDU 3255 farming

Given n Rectangles and M plants, and then given the Weight PI of each plant, PI represents the land of plant I, and Pi can be harvested per unit area, given the coordinates of the lower left corner and the upper right corner of a rectangle, and S, s indicates that the rectangle can contain plants. Ask about the overall maximum benefit.

Solution: because only one plant can be planted, We need to select the most profitable plant for an overlapping land. Aside from this, the rest is the application of the scanning line of the Line Segment tree. Pi can be regarded as the three-dimensional coordinate, and there are only three types of plants, so we can directly discretization. Note that we should discrete the values based on the weights of plant benefits.

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;vector<int> pos;const int maxn = 120000;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)|1)int lc[maxn << 2], rc[maxn << 2], v[maxn << 2], s[maxn << 2];inline void pushup(int u) {    if (v[u])        s[u] = pos[rc[u]+1] - pos[lc[u]];    else if (lc[u] == rc[u])        s[u] = 0;    else        s[u] = s[lson(u)] + s[rson(u)];}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] = s[u] = 0;    if (l == r)        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;    }    friend bool operator < (const Seg& a, const Seg& b) {        return a.x < b.x;    }};typedef long long ll;typedef pair<int,int> pii;int N, M, P[10];pii H[10];vector<Seg> vec[10];inline int find (int k) {    return lower_bound(pos.begin(), pos.end(), k) - pos.begin();}void init () {    scanf("%d%d", &N, &M);    for (int i = 1; i <= M; i++) {        vec[i].clear();        scanf("%d", &H[i].first);        H[i].second = i;    }    sort(H, H + M + 1);    for (int i = 0; i <= M; i++)        P[H[i].second] = i;    int x1, x2, y1, y2, d;    for (int i = 0; i < N; i++) {        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &d);        for (int j = 1; j <= P[d]; j++) {            vec[j].push_back(Seg(x1, y1, y2, 1));            vec[j].push_back(Seg(x2, y1, y2, -1));        }    }}ll solve (int idx) {    pos.clear();    sort(vec[idx].begin(), vec[idx].end());    for (int i = 0; i < vec[idx].size(); i++) {        pos.push_back(vec[idx][i].l);        pos.push_back(vec[idx][i].r);    }    sort(pos.begin(), pos.end());    build(1, 0, pos.size());    ll ret = 0;    for (int i = 0; i < vec[idx].size(); i++) {        modify(1, find(vec[idx][i].l), find(vec[idx][i].r) - 1, vec[idx][i].d);        if (i + 1 != vec[idx].size())            ret += 1LL * s[1] * (vec[idx][i+1].x - vec[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 = 1; i <= M; i++)            ans += 1LL * (H[i].first - H[i-1].first) * solve(i);        printf("Case %d: %I64d\n", kcas, ans);    }    return 0;}

HDU 3255 farming (scanning line)