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
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71

const int N = 305;
const int M = 0;
//Floyd 分别对直飞航班和纪念品价值建图，分别求出前者最小值和满足前者情况下的后者最大值
int num[N][N]; // 表示i到j需经过的最小直飞航班数量
int w[N];
char mp[N][N];
int val[N][N];// 表示i到j在最小直飞航班数量下能购买的最大的纪念品价值
void solve()
{
    int n, q;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> w[i];
    //Floyd初始化
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (i == j)
                num[i][j] = 0;
            else
                num[i][j] = INF;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        cin.ignore();
        for (int j = 1; j <= n; j++)
        {
            cin >> mp[i][j];
            if (mp[i][j] == 'Y')
            {
                num[i][j] = 1;
                val[i][j] = w[j];
            }
        }
    }
	//K要在最外层!
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {	
                //使直飞航班数量最少
                if(num[i][j] > num[i][k] + num[k][j])
                {
                    num[i][j] = num[i][k] + num[k][j];
                    val[i][j] = val[i][k] + val[k][j];
                }
                //在直飞航班数量最少的情况下，使得获得的纪念品价值最大
                else if((num[i][j] == num[i][k] + num[k][j])&&(val[i][j] < val[i][k] + val[k][j]))
                {
                    val[i][j] = val[i][k] + val[k][j];
                }
            }
        }
    }
    cin >> q;
    while (q--)
    {
        int u, v;
        cin >> u >> v;
        if(num[u][v]==INF)
        {
            cout << "Impossible" << endl;
            continue;
        }
        cout << num[u][v] << ' ' << val[u][v]+w[u] << endl;//需加上起点的纪念品价值
    }
}