题意:
求不重叠的矩形个数。
当然,题意是说求能够被扩大的矩形个数,很坑爹,一开始以为扩大的单位是1格,其实题目中明确指出: at least some amount without bumping into another barn,英文是硬伤,所以只要不重叠即可。
思路参考hankcs,把矩形抽象成一个个区间,对应于水平区间和垂直区间,定义每个区间的起点和终点。设置一个计数器,遇到起点计数+1,遇到终点计数-1。
那么矩形不重叠等价于对应于的起点计数不超过2。
接着分别沿x轴扫描,和y轴扫描即可。注意一个细节,排序时,横纵坐标相同时,起点优先,因为corner相碰算重叠。
代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/P3168.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MAX_N = 25000 + 16;
class Point implements Comparable<Point>{
int ord; // 当前轴坐标
int pos; // 另一轴上的坐标
int id; // 对应的矩阵id
int type; // 起点: 0 终点: 1
Point(int ord, int pos, int id, int type){
this.ord = ord;
this.pos = pos;
this.id = id;
this.type = type;
}
@Override
public int compareTo(Point that) {
if (this.ord != that.ord) {
return this.ord - that.ord;
}
if (this.pos != that.pos) {
return this.pos - that.pos;
}
return this.type - that.type;
}
}
int N;
Point[] toX = new Point[4 * MAX_N]; // 以x轴方向扫描
Point[] toY = new Point[4 * MAX_N]; // 以y轴方向扫描
boolean[] ok = new boolean[MAX_N];
void scan(Point[] ps) {
for (int i = 0; i < 4 * N; ++i) {
int j = i;
while (j < 4 * N && ps[j].ord == ps[i].ord) j++;
boolean invalid = false;
int share = 0;
for (int k = i; k < j; ++k) {
Point p = ps[k];
if (invalid) {
ok[p.id] = false;
}
if (p.type == 0) {
share ++;
if (share >= 2) {
invalid = true;
}
}
else {
share --;
if (share == 0) {
invalid = false;
}
}
}
i = j - 1;
}
}
void solve() {
Arrays.sort(toX, 0, 4 * N);
Arrays.sort(toY, 0, 4 * N);
Arrays.fill(ok, true);
scan(toX);
scan(toY);
int ans = 0;
for (int i = 0; i < N; ++i) {
if (ok[i]) ans ++;
}
out.println(ans);
}
void read() {
N = ni();
for (int i = 0; i < N; ++i) {
int A = ni(), B = ni(), C = ni(), D = ni();
toX[4 * i] = new Point(A, B, i, 0);
toX[4 * i + 1] = new Point(A, D, i, 1);
toX[4 * i + 2] = new Point(C, B, i, 0);
toX[4 * i + 3] = new Point(C, D, i, 1);
toY[4 * i] = new Point(B, A, i, 0);
toY[4 * i + 1] = new Point(B, C, i, 1);
toY[4 * i + 2] = new Point(D, A, i, 0);
toY[4 * i + 3] = new Point(D, C, i, 1);
}
solve();
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}