题目

思路：将此坐标转化为直角坐标系（纵坐标，横坐标），然后将六条路径优化，上cs = min(c6, c5 + c1), 下cx = min(c3, c4 + c2), 左cz = min(c5, c4 + c6), 右cy = min(c2, c3 + c1)。因为还有右上、右下的路径，所以干脆一并优化了 cys = min(c1, cy + cs), czs = cz + cs, cyx = cy + cx, czx = min(c4, cz + cx);//cys右上 czs 左上… 然后直接计算即可，详细看代码。

Code:

#include<iostream>
#include<queue>
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;

int main()
{
    
    
    int t;cin >> t;
    while (t--)
    {
    
    
        int x, y;cin >> y >> x;
        ll c1, c2, c3, c4, c5, c6;cin >> c1 >> c2 >> c3 >> c4 >> c5 >> c6;
        ll cs = min(c6, c5 + c1), cx = min(c3, c4 + c2), cz = min(c5, c4 + c6), cy = min(c2, c3 + c1);//cs上  cx下 cz左 cy右
        ll cys = min(c1, cy + cs), czs = cz + cs, cyx = cy + cx, czx = min(c4, cz + cx);//cys右上 .....
        if (x >= 0 && y >= 0)
        {
    
    
            ll c = min(x, y);
            x -= c, y -= c;
            cout << c * cys + x * cy + y * cs << endl;
        }
        else if (x >= 0 && y < 0)
        {
    
    
            y = abs(y);
            ll c = min(x, y);
            x -= c, y -= c;
            cout << c * cyx + x * cy + y * cx << endl;
        }
        else if (x < 0 && y < 0)
        {
    
    
            y = abs(y);x = abs(x);
            ll c = min(x, y);
            x -= c, y -= c;
            cout << c * czx + x * cz + y * cx << endl;
        }
        else if (x < 0 && y >= 0)
        {
    
    
            y = abs(y);x = abs(x);
            ll c = min(x, y);
            x -= c, y -= c;
            cout << c * czs + x * cz + y * cs << endl;
        }
    }
}