uva-1198

题意理解：这道题就是给出了一个图，然后又很多次查询，查询中给出了一些点，如果这些点任意两点之间的最短路径，经过的点都算上。能够凑满整个图那就是yes，否则就是no。

解题思路：我们就是要找最短路，找到任意两点之间的最短路，因为点最多有40个，我们可以用一个 long long 来存看是否经过了所有的点。

Description

Let G = (V,E) be a connected graph without loops and multiple edges, where V and E are the vertex and edge, respectively, sets of G. For any two vertices u, v ∈ V ,the distance between vertices u and v in G is the number of edges in a shortest u-v path. A shortest path between u and v is called a u-v geodesic. Let I(u, v) denote the set of vertices such that a vertex is in I(u, v) if and only if it is in some u-v geodesic of G and, for a set S <= V , I(S) =  ∪ _u,v∈SI(u, v). A vertex set D in graph G is called a geodetic set if I(D) = V . The geodetic set problem is to verify whether D is a geodetic set or not. We use Figure 3 as an example. In Figure 3, I(2, 5) = {2, 3, 4, 5} since there are two shortest paths between vertices 2 and 5. We can see that vertices 3 and 4 are lying on one of these two shortest paths respectively. However, I(2, 5) is not a geodetic set since I(2, 5) != V . Vertex set {1, 2, 3, 4, 5} is intuitively a geodetic set of G. Vertex set D = {1, 2, 5} is also a geodetic set of G since vertex 3 (respectively,vertex 4) is in the shortest path between vertices 1 and 5 (respectively, vertices 2 and 5). Thus, I(D) = V . Besides, vertex sets {1, 3, 4} and {1, 4, 5} are also geodetic sets. 
However, D = {3, 4, 5} is not a geodetic set since vertex 1 is not in I(D).

Input

The input file consists of a given graph and several test cases. The first line contains an integer n indicating the number of vertices in the given graph,where 2 <= n <= 40. The vertices of a graph are labeled from 1 to n. Each vertex has a distinct label. The following n lines represent the adjacent vertices of vertex i, i = 1, 2, ... , n. For example, the second line of the sample input indicates that vertex 1 is adjacent with vertices 2 and 3. Note that any two integers in each line are separated by at least one space. After these n lines, there is a line which contains the 

number of test cases. Each test case is shown in one line and represents a given subset D of vertices. You have to determine whether D is a geodetic set or not. 

Output

For each test case, output 'yes' in one line if it is a geodetic set or 'no' otherwise.

Sample Input

5
2 3
1 3 4
1 2 5
2 5
3 4
6
1 2 3 4 5
1 2 5
2 4
1 3 4
1 4 5
3 4 5

Sample Output

yes
yes
no
yes
yes
no

#include
#include
#include
using namespace std;
#define inf 0x3fffffff
int map_[100][100];
int book2[100];
long long f[100][100];
int main()
{
    int n;
        scanf("%d",&n);
        memset(f,0,sizeof(f));
        for(int i=1;i<50;i++)
        {
            for(int j=1;j<50;j++)
            {
                if(i==j)
                {
                    map_[i][j]=0;
                }
                else
                {
                    map_[i][j]=inf;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            while(1)
            {
                int x;
                scanf("%d",&x);
                map_[i][x]=1;
                char c=getchar();
                if(c=='\n')
                    break;
            }
        }
        for(int k=1;k<=n;k++)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(map_[i][j]>map_[i][k]+map_[k][j])
                    {
                        map_[i][j]=map_[i][k]+map_[k][j];
                    }
                }
            }
        }
        for(int k=1;k<=n;k++)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(map_[i][j]==map_[i][k]+map_[k][j])
                    {
                        f[i][j]|=((long long)1<<(long long)(k-1));
                    }
                }
            }
        }
        int q;
        scanf("%d",&q);
        for(int i=0;i>1;
            }
            if(cnt==n)
            {
                printf("yes\n");;
            }
            else
            {
                printf("no\n");
            }
        }
    }