求更优解

题目

我的代码：

#include <bits/stdc++.h>
using namespace std;
int n,x,f[305][10],ans[305][10];
int dx[8] = {1,1,1,0,0,-1,-1,-1};
int dy[8] = {-1,0,1,-1,1,-1,0,1};
struct node{
    int x,y,step;
};
queue <node> q;
void paint (int c,int s){
    if (s == 0){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j++){
                if (i == 0 || i == 4 || j == 1 || j == 5){
                    f[c * 5 - i][j] = 1;
                }
            }
        }
    }else if (s == 1){
        for (int i = 0 ; i < 5 ; i++){
            f[c * 5 - i][5] = 1;
        }
        for (int i = 1 ; i <= 5 ; i++){
            f[c * 5 - 2][i] = 1;
        }
        f[c * 5 - 3][2] = 1;
        f[c * 5 - 4][3] = 1;
    }else if (s == 2){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j++){
                if (!(j == 2 && i != 0 || j == 4 && i != 4)){
                    f[c * 5 - i][j] = 1;
                }
            }
        }
    }else if (s == 3){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j++){
                if (!((j == 2 || j == 4) &&  i != 0)){
                    f[c * 5 - i][j] = 1;
                }
            }
        }
    }else if (s == 4){
        for (int i = 1 ; i <= 5 ; i++){
            f[c * 5][i] = 1;
        }
        for (int i = 1 ; i <= 3 ; i++){
            f[c * 5 - 4][i] = 1;
        }
        for (int i = 1 ; i <= 3 ; i++){
            f[c * 5 - i][3] = 1;
        }
    }else if (s == 5){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j++){
                if (!(j == 4 && i != 0 || j == 2 && i != 4)){
                    f[c * 5 - i][j] = 1;
                }
            }
        }
    }else if (s == 6){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j += 2){
                f[c * 5 - i][j] = 1;
            }
        }
        f[c * 5][2] = 1;
        f[c * 5 - 4][4] = 1;
        f[c * 5][4] = 1;
    }else if (s == 7){
        for (int i = 0 ; i < 5 ; i++){
            f[c * 5 - i][1] = 1;
        }
        for (int i = 1 ; i <= 5 ; i++){
            f[c * 5][i] = 1;
        }
    }else if (s == 8){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j += 2){
                f[c * 5 - i][j] = 1;
            }
        }
        for (int i = 1 ; i <= 5 ; i++){
            f[c * 5 - 4][i] = 1;
        }
        for (int i = 1 ; i <= 5 ; i++){
            f[c * 5][i] = 1;
        }
    }else if (s == 9){
        for (int i = 0 ; i < 5 ; i++){
            for (int j = 1 ; j <= 5 ; j += 2){
                f[c * 5 - i][j] = 1;
            }
        }
        f[c * 5 - 4][2] = 1;
        f[c * 5][2] = 1;
        f[c * 5][4] = 1;
    }
}
void bfs (){
    q.push((node){1,1,0});
    ans[1][1] = 0;
    while (!q.empty()){
        node tmp = q.front();
        q.pop();
        for (int i = 0 ; i < 8 ; i++){
            int nx = tmp.x + dx[i];
            int ny = tmp.y + dy[i];
            if (nx < 1 || nx > (n * 5) || ny < 1 || ny > 5 || f[nx][ny] == 0 || ans[nx][ny] <= 1000){
                continue;
            }
            q.push((node){nx,ny,tmp.step + 1});
            ans[nx][ny] = tmp.step + 1;
        }
    }
}
int main(){
    cin >> n;
    for (int i = 1 ; i <= n ; i++){
        cin >> x;
        paint(i,x);
    }
    memset (ans,127,sizeof(ans));
    bfs ();
    if (ans[n * 5][5] <= 1e6) cout << ans[n * 5][5];
    else cout << -1;
    return 0;
}