Ultraviolet A-11983 Weird Advertisement (intersection area of a line segment tree)

Ultraviolet A-11983 Weird Advertisement (intersection area of a line segment tree)

Description

G	Weird Advertisement

Renat Mullakhanov (rem), one of the most talented programmers in the world, passed away on March 11,201 1. this is very sad news for all of us. his team went to acm icpc World Finals-2004, placed 4th and won gold medals. he really was a great programmer. may he rest in peace. this problem is dedicated to him.

2 DPlaneLandIs a land just like a huge2DPlane. The rangeXAxis is0To109And the rangeYAxis is also0To109. People built houses only in integer co-ordinates and there is exactly one house in each integer co-ordinate.

NowUseAndSmileSoap Company is launching a new soap. That's why they want to advertise this product as much as possible. So, they selectedNPersons for this task. Each person will be given a rectangular region. He will advertise the product to all the houses that lie in his region. Each rectangular region is identified4IntegersX1, y1, x2AndY2. That means this person will advertise in all the houses whoseXCo-ordinate isX1AndX2(Aggressive) andYCo-ordinate isY1AndY2(Aggressive ).

Now after a while they realized that some houses are being advertised by more than one person. So, they want to find the number of houses that are advertised by at leastKPersons. Since you are one of the best programmers in the city; they asked you to solve this problem.

Input

Input starts with an integerT (≤ 13), Denoting the number of test cases.

Each case starts with a line containing two integersN (1 ≤ n ≤ 30000), k (1 ≤ k ≤ 10). Each of the nextNLines will contain4IntegersX1, y1, x2, y2 (0 ≤ x1, y1, x2, y2 ≤ 109, x1 <x2, y1 <y2)Denoting a rectangular region for a person.

Output

For each case, print the case number and the total number of houses that are advertised by at leastKPeople.

Sample Input	Output for Sample Input
2 2 1 0 0 4 4 1 1 2 5 2 2 0 0 4 4 1 1 2 5	Case 1: 27 Case 2: 8

Here are n rectangles for you to overwrite the number of points greater than or equal to k.

Idea: Convert the vertices x and y in the upper right corner to the sum-up area, or mark 1 on the top, Mark-1 on the bottom, and merge area on the traditional line tree, it is easy to make a mistake to reference Hu Hao when updating a line segment: a node of the Line Segment tree is not an endpoint of the line segment, but a line segment between the endpoint and the next endpoint, so the title of r + 1, R-1 place can be a good bit of their own

#include 
 
  #include 
  
   #include #include 
   
    #define lson(x) ((x) << 1)#define rson(x) ((x) << 1 | 1)#define ll long longusing namespace std;const int maxn = 60005;const int M = 160010;int k;struct Seg {int l, r, h, s;Seg() {}Seg(int a, int b, int c, int d) {l = a, r = b, h = c, s = d;}bool operator <(const Seg &tmp) const {if (h == tmp.h)return s > tmp.s;return h < tmp.h;}} p[maxn<<2];int sum[11][M<<2];int cover[M<<2];int w[M<<2];;void pushup(int pos, int l, int r) {for (int i = 0; i <= k; i++)sum[i][pos] = 0;if (cover[pos] >= k) {sum[k][pos] = w[r+1] - w[l];return;}if (r == l) {sum[cover[pos]][pos] = w[r+1] - w[l];return;}int tmp = 0;for (int i = 0; i <= k; i++) {if (i + cover[pos] < k)sum[i+cover[pos]][pos] = sum[i][lson(pos)] + sum[i][rson(pos)];else sum[k][pos] += sum[i][lson(pos)] + sum[i][rson(pos)];}for (int i = 1; i <= k; i++)tmp += sum[i][pos];sum[0][pos] = w[r+1] - w[l] - tmp;}void update(int L, int R, int l, int r, int pos, int op) {if (L <= l && R >= r) {cover[pos] += op;pushup(pos, l, r);return;}int m = l + r >> 1;if (L <= m)update(L, R, l, m, lson(pos), op);if (R > m)update(L, R, m+1, r, rson(pos), op);pushup(pos, l, r);}int search(int x, int r) {int l = 0;while (l <= r) {int m = l + r >> 1;if (w[m] == x)return m;else if (w[m] < x)l  = m + 1;else r = m - 1;}return -1;}void build(int l, int r, int pos) {sum[0][pos] = w[r+1] - w[l];if (l == r)return;int m = l + r >> 1;build(l, m, lson(pos));build(m+1, r, rson(pos));}int main() {int t, n, m, cas = 1;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &k);memset(cover, 0, sizeof(cover));memset(sum, 0, sizeof(sum));int tot = 0;int a, b, c, d;for (int i = 0; i < n; i++) {scanf("%d%d%d%d", &a, &b, &c, &d);c++, d++;p[tot] = Seg(a, c, b, 1);w[tot++] = a;p[tot] = Seg(a, c, d, -1);w[tot++] = c;}if (k > n) {printf("Case %d: 0\n", cas++);continue;}sort(w, w+tot);sort(p, p+tot);int cnt = unique(w, w+tot) - w;build(0, cnt-1, 1);ll ans = 0;for (int i = 0; i < tot-1; i++) {int l = search(p[i].l, cnt-1);int r = search(p[i].r, cnt-1)-1;update(l, r, 0, cnt-1, 1, p[i].s);ans += (ll)(sum[k][1] * (ll)(p[i+1].h - p[i].h));}printf("Case %d: ", cas++);printf("%lld\n", ans);}return 0;}