알고리즘
백준 15685번 드래곤 커브 C++
영춘권의달인
2022. 7. 31. 10:58
어떤 점을 중심으로 시계방향으로 90도 회전했을 때의 좌표를 구하는 방법만 생각해낸다면 쉽게 풀 수 있는 문제이다.
#include <iostream>
#include <string>
#include <vector>
#include <math.h>
#include <algorithm>
#include <utility>
#include <stack>
#include <queue>
#include <math.h>
#include <set>
#include <map>
#include <list>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <limits.h>
using namespace std;
using int64 = long long;
int dX[4] = { 1,0,-1,0 };
int dY[4] = { 0,-1,0,1 };
vector<vector<bool>> board(101, vector<bool>(101, false));
vector<vector<pair<int, int>>> dragons;
vector<bool> isValid;
vector<int> currGeneration;
vector<int> targetGeneration;
pair<int, int> convertPos(pair<int, int> ¢or, pair<int, int> &pos) {
int x = pos.first - centor.first;
int y = pos.second - centor.second;
::swap(x, y);
x = -x;
x += centor.first;
y += centor.second;
return ::make_pair(x, y);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
dragons = vector<vector<pair<int, int>>>(n, vector<pair<int, int>>());
isValid = vector<bool>(n, true);
currGeneration = vector<int>(n, 0);
targetGeneration = vector<int>(n, 0);
for (int i = 0; i < n; i++) {
int x, y, d, g;
cin >> x >> y >> d >> g;
dragons[i].push_back({ x,y });
board[y][x] = true;
int nextX = x + dX[d];
int nextY = y + dY[d];
if (nextX < 0 || nextX>100) continue;
if (nextY < 0 || nextY>100) continue;
dragons[i].push_back({ nextX,nextY });
board[nextY][nextX] = true;
targetGeneration[i] = g;
}
for (int i = 0; i < n; i++) {
while (currGeneration[i] < targetGeneration[i]) {
int lastIdx = dragons[i].size() - 1;
pair<int, int> centor = dragons[i][lastIdx];
vector<pair<int, int>> addPoses;
for (int j = lastIdx - 1; j >= 0; j--) {
addPoses.push_back(convertPos(centor, dragons[i][j]));
}
for (auto a : addPoses) {
dragons[i].push_back(a);
board[a.second][a.first] = true;
}
currGeneration[i]++;
}
}
int ans = 0;
for (int i = 0; i < 100; i++) {
for (int j = 0; j < 100; j++) {
bool check1 = board[i][j];
bool check2 = board[i + 1][j];
bool check3 = board[i][j + 1];
bool check4 = board[i + 1][j + 1];
if (check1 && check2 && check3 && check4)
ans++;
}
}
cout << ans;
}