티스토리 뷰
문제 : https://www.acmicpc.net/problem/2983
1. x+y 와 x-y 의 케이스로 나눠 각각의 케이스 탐색
2. 시간초과로 실패...
소스코드
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
/*
*/
public class Baekjoon2983 {
//class Main {
static double initX = 0;
static double initY = 0;
public static void main(String[] args)
throws IOException
{
// long start = System.currentTimeMillis();
FileReader fr = new FileReader("D:\\dev\\worksspace\\Solution\\src\\Baekjoon2983");
// InputStreamReader fr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(fr);
String s = null;
while ((s=br.readLine())!=null) {
int N = Integer.parseInt(s.split(" ")[0]);
int K = Integer.parseInt(s.split(" ")[1]);
String jumpStr = br.readLine();
String[] jumbIndex = null;
double X;
double Y;
double xySumTemp;
double xMinusYTemp;
HashMap<Double, ArrayList<Double>> xySum = new HashMap<Double, ArrayList<Double>>();
HashMap<Double, ArrayList<Double>> xMinusY = new HashMap<Double, ArrayList<Double>>();
initX = 0;
initY = 0;
for (int i = 0; i < N; i++) {
jumbIndex = br.readLine().split(" ");
X = Double.parseDouble(jumbIndex[0]);
Y = Double.parseDouble(jumbIndex[1]);
if(initX == 0) initX = X;
if(initY == 0) initY = Y;
xySumTemp = X+Y;
xMinusYTemp = X-Y;
if(xySum.get(xySumTemp) == null){
ArrayList<Double> tempList = new ArrayList<Double>();
tempList.add(X);
xySum.put(xySumTemp, tempList);
}else{
xySum.get(xySumTemp).add(X);
}
if(xMinusY.get(xMinusYTemp) == null){
ArrayList<Double> tempList = new ArrayList<Double>();
tempList.add(X);
xMinusY.put(xMinusYTemp, tempList);
}else{
xMinusY.get(xMinusYTemp).add(X);
}
}
String gubun = null;
for (int i = 0; i < jumpStr.length(); i++) {
double nextX = 0;
double nextY = 0;
gubun = jumpStr.substring(i, i+1);
int index = -1;
double tempXMinusY = initX-initY;
double tempXPlusY = initX+initY;
double diffWithX = 1000000001;
boolean tb;
if("A".equals(gubun)){
ArrayList<Double> tempList = xMinusY.get(tempXMinusY);
for (int j = 0; j < tempList.size(); j++) {
double tempJ = tempList.get(j);
if(tempJ > initX && (tempJ-initX < diffWithX)){
nextX = tempJ;
nextY = tempJ-tempXMinusY;
diffWithX = tempJ-initX;
index = j;
}
}
if(nextX != 0){
tb = xMinusY.get(tempXMinusY).remove(initX);
tb = xySum.get(tempXPlusY).remove(initX);
initX = nextX;
initY = nextY;
}
}else if("D".equals(gubun)){
ArrayList<Double> tempList = xMinusY.get(tempXMinusY);
for (int j = 0; j < tempList.size(); j++) {
double tempJ = tempList.get(j);
if(tempJ < initX && (initX-tempJ < diffWithX)){
nextX = tempJ;
nextY = tempJ-tempXMinusY;
diffWithX = initX-tempJ;
index = j;
}
}
if(nextX != 0){
tb = xMinusY.get(tempXMinusY).remove(initX);
tb = xySum.get(tempXPlusY).remove(initX);
initX = nextX;
initY = nextY;
}
}else if("B".equals(gubun)){ //(x+P, y-P)
ArrayList<Double> tempList = xySum.get(tempXPlusY);
for (int j = 0; j < tempList.size(); j++) {
double tempJ = tempList.get(j);
if(tempJ > initX && (tempJ-initX < diffWithX)){
nextX = tempJ;
nextY = tempXPlusY-tempJ;
diffWithX = tempJ-initX;
index = j;
}
}
if(nextX != 0){
tb = xMinusY.get(tempXMinusY).remove(initX);
tb = xySum.get(tempXPlusY).remove(initX);
initX = nextX;
initY = nextY;
}
}else if("C".equals(gubun)){ //(x-P, y+P)
ArrayList<Double> tempList = xySum.get(tempXPlusY);
for (int j = 0; j < tempList.size(); j++) {
double tempJ = tempList.get(j);
if(initX > tempJ && (initX-tempJ < diffWithX)){
nextX = tempJ;
nextY = tempXPlusY-tempJ;
diffWithX = initX-tempJ;
index = j;
}
}
if(nextX != 0){
tb = xMinusY.get(tempXMinusY).remove(initX);
tb = xySum.get(tempXPlusY).remove(initX);
initX = nextX;
initY = nextY;
}
}
}
}
System.out.println((int)initX+" "+(int)initY);
// long end = System.currentTimeMillis();
// System.out.println("time "+ (end-start)/1000.0);
}
}
'Programming > Algorithm' 카테고리의 다른 글
[Algorithm 1일차] 막대기 자르기 (0) | 2016.11.14 |
---|---|
[Algorithm 1일차] Assembly Line Scheduling (0) | 2016.11.14 |
[Segment Tree] 백준 알고리즘 2042번 구간 합 (1329) | 2016.11.03 |
[중상] 서로소 (578) | 2016.10.31 |
[중상] 집합 (1043) | 2016.10.31 |
댓글
공지사항
최근에 올라온 글
최근에 달린 댓글
- Total
- Today
- Yesterday
링크
TAG
- 중국어공부
- GraphQL
- 서머너즈워
- 자금조달계획서
- 중국어강의
- 중국어정리
- 노브랜드
- s9+
- 마시내 탕수육
- 로꼬
- AWS nodejs
- 알고리즘
- 크러쉬
- 부동신 계약시 주의사항
- AWS npm
- 프리티어
- Bitnami
- Git
- 프렌즈
- 웰빙헬스
- 10cm
- 부동산거래계약신고필증
- 혁오
- 고운발크림
- 존맛탱
- 뒤꿈치 건조함
- Axis2
- ES6
- 수미네 반찬
- 생활코딩
일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | 2 | |||||
3 | 4 | 5 | 6 | 7 | 8 | 9 |
10 | 11 | 12 | 13 | 14 | 15 | 16 |
17 | 18 | 19 | 20 | 21 | 22 | 23 |
24 | 25 | 26 | 27 | 28 | 29 | 30 |
31 |
글 보관함