Adding New Machine
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1555 Accepted Submission(s): 353
Problem Description
Incredible Crazily Progressing Company (ICPC) suffered a lot with the low speed of procedure. After investigation, they found that the bottleneck was at Absolutely Crowded Manufactory (ACM). In oder to accelerate the procedure, they bought a new machine for ACM. But a new problem comes, how to place the new machine into ACM?
ACM is a rectangular factor and can be divided into W * H cells. There are N retangular old machines in ACM and the new machine can not occupy any cell where there is old machines. The new machine needs M consecutive cells. Consecutive cells means some adjacent cells in a line. You are asked to calculate the number of ways to choose the place for the new machine.
Input
There are multiple test cases (no more than 50). The first line of each test case contains 4 integers W, H, N, M (1 ≤ W, H ≤ 107, 0 ≤ N ≤ 50000, 1 ≤ M ≤ 1000), indicating the width and the length of the room, the number of old machines and the size of the new machine. Then N lines follow, each of which contains 4 integers Xi1, Yi1, Xi2 and Yi2 (1 ≤ Xi1 ≤ Xi2 ≤ W, 1 ≤ Yi1 ≤ Yi2 ≤ H), indicating the coordinates of the i-th old machine. It is guarantees that no cell is occupied by two old machines.
Output
Output the number of ways to choose the cells to place the new machine in one line.
Sample Input
3 3 1 2
2 2 2 2
3 3 1 3
2 2 2 2
2 3 2 2
1 1 1 1
2 3 2 3
Sample Output
8
4
3
题解
此题一眼扫描线。
我们可以先算出所有的方案，之后算出线段的下（右）段在矩阵的方案数，一减就是答案。
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 100005
struct Line {
int lx, rx, h, flag;
Line(int lx, int rx, int h, int flag) :
lx(lx), rx(rx), h(h), flag(flag) {}
bool operator < (const Line &a) const {
return h < a.h;
}
};
struct Point {
int x1, y1, x2, y2;
} p[N];
struct segment {
int l, r, flag;
long long sum;
}C[N << 2];
int n;
long long ans;
vector<int> a;
vector<Line> line;
int idx(const int &x) {
return lower_bound(a.begin(), a.end(), x) - a.begin();
}
#define ls u << 1
#define rs u << 1 | 1
void pu(int u) {
if (C[u].flag)
C[u].sum = a[C[u].r] - a[C[u].l];
else if (C[u].l + 1 == C[u].r)
C[u].sum = 0;
else
C[u].sum = C[ls].sum + C[rs].sum;
}
void build(int u, int l, int r) {
C[u] = (segment){l, r, 0, 0LL};
if (l + 1 == r)
return;
int mid = (l + r) >> 1;
build(ls, l, mid);
build(rs, mid, r);
}
void upd(int u, const int &x, const int &y, const int &v) {
if (x <= C[u].l && C[u].r <= y) {
C[u].flag += v;
pu(u);
return;
}
int mid = (C[u].l + C[u].r) >> 1;
if (x < mid) upd(ls, x, y, v);
if (mid < y) upd(rs, x, y, v);
pu(u);
}
void solve(bool flag, int w, int h, int m) {
a.clear(), line.clear();
a.push_back(1);
a.push_back(w + 1);
int tmp = max(h - m + 2, 1);
line.push_back(Line(1, w + 1, tmp, 1));
line.push_back(Line(1, w + 1, h + 1, -1));
for (int i = 1; i <= n; ++i) {
if (!flag) {
scanf("%d%d%d%d", &p[i].x1, &p[i].y1, &p[i].x2, &p[i].y2);
++p[i].x2, ++p[i].y2;
} else {
swap(p[i].x1, p[i].y1);
swap(p[i].x2, p[i].y2);
}
tmp = max(p[i].y1 - m + 1, 1);
line.push_back(Line(p[i].x1, p[i].x2, tmp, 1));
line.push_back(Line(p[i].x1, p[i].x2, p[i].y2, -1));
a.push_back(p[i].x1);
a.push_back(p[i].x2);
}
sort(a.begin(), a.end());
sort(line.begin(), line.end());
a.erase(unique(a.begin(), a.end()), a.end());
int sza = a.size(), szline = line.size();
build(1, 0, sza);
for (int i = 0; i < szline; ++i) {
if (i)
ans -= C[1].sum * (long long)(line[i].h - line[i - 1].h);
upd(1, idx(line[i].lx), idx(line[i].rx), line[i].flag);
}
}
int main() {
int w, h, m;
while (scanf("%d%d%d%d", &w, &h, &n, &m) == 4) {
ans = 2LL * w * h;
solve(false, w, h, m);
solve(true, h, w, m);
if (m == 1)
ans >>= 1;
printf("%lld
", ans);
}
return 0;
}