Ultraviolet A 1492-Adding New Machine (line segment tree)

Ultraviolet A 1492-Adding New Machine (line segment tree)

Link to the question: Ultraviolet A 1492-Adding New Machine

In an R? There are N old machines in the C matrix. Given the land occupation of each machine, now we need to add 1? M machine, ask how many placement methods are there.

Problem-solving ideas: the problem can be converted into a rectangular coverage problem, for each old machine, assuming that the corresponding position of the right to put, then the left of the M-1 is not allowed to put, and the position of the M-1 on the left of the right boundary. Use the line segment tree to solve rectangular overwrite, and the x and y coordinates are processed once. Pay attention to M = 1.

#include 
  
   #include 
   
    #include 
    
     #include using namespace std;typedef long long ll;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)+1)const int maxn = 50100;ll N;int M;int x[2][maxn], y[2][maxn];vector
     
       bset, vec;struct Segment {    int l, r, h, v;    void set(int l, int r, int h, int v) {        this->l = l;        this->r = r;        this->h = h;        this->v = v;    }}seg[maxn*2], node[maxn*8];inline bool cmp (const Segment& a, const Segment& b) {    return a.h < b.h;}inline int search (int x) {    return lower_bound(bset.begin(), bset.end(), x) - bset.begin();}inline void get_hashkey () {    sort(vec.begin(), vec.end());    bset.push_back(vec[0]);    for (int i = 1; i < vec.size(); i++) {        if (vec[i] != vec[i-1])            bset.push_back(vec[i]);    }}inline void add_segment (int i, int h, int l, int r, int v) {    seg[i].set(l, r, h, v);    vec.push_back(l);    vec.push_back(r);}inline void pushup (int u) {    if (node[u].h)        node[u].v = bset[node[u].r+1] - bset[node[u].l];    else if (node[u].l == node[u].r)        node[u].v = 0;    else        node[u].v = node[lson(u)].v + node[rson(u)].v;}void build_segTree (int u, int l, int r) {    node[u].set(l, r, 0, 0);    if (l == r) {        pushup(u);        return ;    }    int mid = (l + r) / 2;    build_segTree(lson(u), l, mid);    build_segTree(rson(u), mid + 1, r);    pushup(u);}ll query_segTree (int u, int l, int r) {    if (l <= node[u].l && node[u].r <= r)        return node[u].v;    int mid = (node[u].l + node[u].r) / 2;    ll ret = 0;    if (l <= mid)        ret += query_segTree(lson(u), l, r);    if (r > mid)        ret += query_segTree(rson(u), l, r);    return ret;}void add_segTree (int u, int l, int r, int v) {    if (l <= node[u].l && node[u].r <= r) {        node[u].h += v;        pushup(u);        return;    }    int mid = (node[u].l + node[u].r) / 2;    if (l <= mid)        add_segTree(lson(u), l, r, v);    if (r > mid)        add_segTree(rson(u), l, r, v);    pushup(u);}ll solve (int x[2][maxn], int y[2][maxn], ll R, ll C) {    if (N == 0)        return 1LL * C * (R - M + 1);    int n = 0;    vec.clear();    bset.clear();    for (int i = 0; i < N; i++) {        add_segment(n++, x[0][i] - 1, max(y[0][i] - M, 0), y[1][i], 1);        add_segment(n++, x[1][i], max(y[0][i] - M, 0), y[1][i], -1);    }    add_segment(n++, 0, max((int)R + 1 - M, 0), R, 1);    add_segment(n++, C, max((int)R + 1 - M, 0), R, -1);    get_hashkey();    sort(seg, seg + n, cmp);    int size = bset.size() - 2;    build_segTree (1, 0, size);    ll ret = 0;    for (int i = 0; i < n; i++) {        ll tmp = query_segTree(1, 0, size);        if (i)            ret += tmp * (seg[i].h - seg[i-1].h);        add_segTree(1, search(seg[i].l), search(seg[i].r) - 1, seg[i].v);    }    return R * C - ret;}int main () {    ll R, C;    while (scanf("%lld%lld%lld%d", &R, &C, &N, &M) == 4) {        for (int i = 0; i < N; i++)            scanf("%d%d%d%d", &x[0][i], &y[0][i], &x[1][i], &y[1][i]);        ll ans = solve(x, y, C, R) + solve(y, x, R, C);        if (M == 1)            ans /= 2;        printf("%lld\n", ans);    }    return 0;}