UVA - 11983 Weird Advertisement （线段树求并面积）

Description

G

Weird Advertisement

Renat Mullakhanov (rem), one of the most talented programmers in the world, passed away on March 11, 2011. 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.

2DPlaneLand is a land just like a huge 2D plane. The range ofX axis is 0 to 10⁹ and the range ofY axis is also 0 to 10⁹. People built houses only in integer co-ordinates and there is exactly one house in each integer co-ordinate.

Now UseAndSmile Soap Company is launching a new soap. That's why they want to advertise this product as much as possible. So, they selectedn persons 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 identified by4 integers x₁, y₁, x₂ andy₂. That means this person will advertise in all the houses whosex co-ordinate is between x₁ and x₂ (inclusive) and y co-ordinate is between y₁ and y₂ (inclusive).

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 leastk persons. Since you are one of the best programmers in the city; they asked you to solve this problem.

Input

Input starts with an integer T (≤ 13), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 30000), k (1 ≤ k ≤ 10). Each of the nextn lines will contain 4 integers x₁, 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 leastk people.

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

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题意：给你n个矩形，求覆盖大于等于k的点数

思路：将右上角的点x，y都加一就能转换成求并面积了，还是上面的边标记1，下面的边标记-1，传统的线段树求并面积，容易错的地方是在更新的时候，引用胡浩大大的一句话：这里线段树的一个结点并非是线段的一个端点,而是该端点和下一个端点间的线段,所以题目中r+1,r-1的地方可以自己好好的琢磨一下

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#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;}