hdu 2807 The Shortest Path（矩阵+floyd）

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3164    Accepted Submission(s):
1030

Problem Description

There are N cities in the country. Each city is
represent by a matrix size of M*M. If city A, B and C satisfy that A*B = C, we
say that there is a road from A to C with distance 1 (but that does not means
there is a road from C to A).
Now the king of the country wants to ask me
some problems, in the format:
Is there is a road from city X to Y?
I have
to answer the questions quickly, can you help me?

 

Input

Each test case contains a single integer N, M,
indicating the number of cities in the country and the size of each city. The
next following N blocks each block stands for a matrix size of M*M. Then a
integer K means the number of questions the king will ask, the following K lines
each contains two integers X, Y(1-based).The input is terminated by a set
starting with N = M = 0. All integers are in the range [0, 80].

 

Output

For each test case, you should output one line for each
question the king asked, if there is a road from city X to Y? Output the
shortest distance from X to Y. If not, output "Sorry".

 

Sample Input

3 2

1 1

2 2

1 1

1 1

2 2

4 4

1

1 3

3 2

1 1

2 2

1 1

1 1

2 2

4 3

1

1 3

0 0

 

Sample Output

1
Sorry

 

Source

HDU
2009-4 Programming Contest

 

题意：先输入n，m，接着输入n个m*m的矩阵，再输入k表示k次询问，接下来输入k行输入x，y，询问x点是否能到y点，可以则输出最短路。 当矩阵 A*B = C 的时候，A点到C点的距离是1.

 

先处理矩阵，建图，此题数据只有85,直接用floyd就好。

 

附上代码：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define N 85
 #define INF 0x3f3f3f3f
 using namespace std;
 int n,m;
 int map[N][N];
 int a[N][N][N],tem[N][N];

 void floyd()
 {
     int i,j,k;
     for(k=; k<=n; k++)
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
                 if(map[i][j]>map[i][k]+map[k][j])
                     map[i][j]=map[i][k]+map[k][j];
 }

 void getmap()
 {
     int i,j,k,x,y,z;
     for(i=; i<=n; i++)
         for(j=; j<=m; j++)
             for(k=; k<=m; k++)
                 scanf("%d",&a[i][j][k]);
     for(i=; i<=n; i++)
         for(j=; j<=n; j++)
             if(i==j) map[i][j]=;
             else   map[i][j]=INF;
     for(i=; i<=n; i++)
     {
         for(j=; j<=n; j++)
         {
             if(i==j) continue;
             memset(tem,,sizeof(tem));
             for(x=; x<=m; x++)
                 for(y=; y<=m; y++)
                 {
                     tem[x][y]=;
                     for(z=; z<=m; z++)   ///矩阵计算
                         tem[x][y]+=a[i][x][z]*a[j][z][y];
                 }
             for(x=; x<=n; x++)
             {
                 if(x==i||x==j)
                     continue;
                 int flag=;
                 for(y=; y<=m; y++)
                 {
                     for(z=; z<=m; z++)
                     {
                         if(tem[y][z]!=a[x][y][z]) ///比较是否完全相同
                         {
                             flag=;
                             break;
                         }
                     }
                     if(!flag)
                         break;
                 }
                 if(flag)
                     map[i][x]=;
             }
         }
     }
     floyd();
 }
 int main()
 {
     int i,j,k,x,y;
     while(~scanf("%d%d",&n,&m))
     {
         if(n==&&m==)
             break;
         getmap();
         scanf("%d",&k);
         while(k--)
         {
             scanf("%d%d",&x,&y);
             if(map[x][y]<INF)
                 printf("%d\n",map[x][y]);
             else
                 printf("Sorry\n");
         }
     }
     return ;
 }