Description
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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 of
X axis is 0 to 109 and the range of
Y 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 selected
n 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 by
4 integers x1, y1, x2 and
y2. That means this person will advertise in all the houses whose
x 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 least
k 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 next
n 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 least
k 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 |
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 long
using 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;
}
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