백준 20057 마법상어와 토네이도(c++)

 

#include<iostream>
#include<vector> 
#include<cmath>
using namespace std;

vector<vector<int>>A;

int curr, curc;

int dr[4] = { 0,1,0,-1 };
int dc[4] = { -1,0,1,0 };
//격자 밖으로 나간 모래의 양
int ans = 0;
int N;
int dx[4][10] = { {-1,1,-2,2,-1,1,-1,1,0,0},{0,0,0,0,-1,-1,1,1,2,1},{-1,1,2,-2,1,-1,1,-1,0,0},{0,0,0,0,1,1,-1,-1,-2,-1} };
int dy[4][10] = { {0,0,0,0,1,1,-1,-1,-2,-1},{-1,1,-2,2,-1,1,-1,1,0,0} ,{0,0,0,0,-1,-1,1,1,2,1} ,{1,-1,2,-2,1,-1,-1,1,0,0} };
double degree[9] = { 0.07,0.07,0.02,0.02,0.01,0.01,0.1,0.1,0.05 }; //double!!!


void fly_away(int nr, int nc, int d) {

    //nr,nc로 이동할것임 

    int total = A[nr][nc];//날아갈 모래의 양 
    
    A[nr][nc] = 0;

    //d방향에 따라 사방으로 날리기 
    //a구하기 위한 날라간 양, 격자 밖으로 날라간 양 
    int tmp = 0; int out = 0;
    //우선 9개 방향만 하기 
    for (int i = 0; i < 9; i++) {
        int gr = nr + dx[d][i];
        int gc = nc + dy[d][i];
        int sub = (total * degree[i]);
        tmp += sub;
        //격자 밖이면 out++;
        if (gr<1 || gr>N || gc<1 || gc>N) {
            out += sub;
            continue;
        }
        //격자 안이면 A에 더하기 
        A[gr][gc] += sub;
      
    }
    //a의 양 
    int a = (total - tmp);

    int ar = nr + dx[d][9];
    int ac = nc + dy[d][9];

    if (ar<1 || ar>N || ac < 1 || ac>N){
        out+=a;
    }
    else {
        A[ar][ac]+=a;
    }
    ans += out;
}


int main() {

    A.assign(505, vector<int>(505, 0));
   
    cin >> N;
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= N; j++) {
            cin >> A[i][j];
        }
    }
    //초기 위치 
    curr = (N + 1) / 2; curc = (N + 1) / 2;

    int cnt = 0;
    int d = 0; //초기 방향 
    bool flag = false;
    int k = 1;
    while (1) {

        cnt++;
       
        //k=1,2,3,4...
        for (int j = 0; j < 2; j++) {
            for (int i = 0; i < k; i++) {

                //다음 위치 
                int nr = curr + dr[d];
                int nc = curc + dc[d];
                
                //모래 흩날리기 
                fly_away(nr, nc, d);
                if (nr == 1 && nc == 1) {
                    flag = true;
                    break;
                }
                curr = nr; curc = nc; //갱신하기
            }
            if (flag)break;
             
            d = (d + 1) % 4;
        }
        if (flag)break;
        cnt = 0;
        k++;
    }
    
    cout << ans;

    return 0;
}

https://www.acmicpc.net/problem/20057

20057번: 마법사 상어와 토네이도

double 배열 선언 조심하자..