hdu 2807 The Shortest Path（矩阵）

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3006    Accepted Submission(s): 987

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 21 12 21 11 12 24 411 33 21 12 21 11 12 24 311 30 0

 

Sample Output

1Sorry

 

Source

HDU 2009-4 Programming Contest

 

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题解：根据题意乱搞即可。。。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 83#define inf 1000000000using namespace std;int n,m,dis[N][N];struct data{int a[N][N];}point[N];data get(data a,data b){data c;for (int i=1;i<=m;i++) for (int j=1;j<=m;j++)  {  c.a[i][j]=0;  for (int k=1;k<=m;k++)   c.a[i][j]=c.a[i][j]+a.a[i][k]*b.a[k][j];  }return c;}bool pd(data a,data b){for (int i=1;i<=m;i++) for (int j=1;j<=m;j++)  if (a.a[i][j]!=b.a[i][j])  return 0;return 1;}void calc(){for (int a=1;a<=n;a++) for  (int b=1;b<=n;b++)  if (a!=b)  {  data t=get(point[a],point[b]);  for (int c=1;c<=n;c++)   {   if (c==a||c==b) continue;   if  (pd(point[c],t))  dis[a][c]=1;   }  }}void floyed(){for (int k=1;k<=n;k++) for (int i=1;i<=n;i++)  if (dis[i][k]!=inf)  for (int j=1;j<=n;j++)   if (dis[k][j]!=inf)    if (k!=i&&i!=j&&j!=k)     dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);}int main(){while (scanf("%d%d",&n,&m)!=EOF){if (!n&&!m)  break;for (int i=1;i<=n;i++) { for (int j=1;j<=m;j++)  for (int k=1;k<=m;k++)   scanf("%d",&point[i].a[j][k]); }for (int i=1;i<=n;i++) for (int j=1;j<=n;j++)  dis[i][j]=inf;calc();floyed();int t; scanf("%d",&t);for (int i=1;i<=t;i++){int x,y; scanf("%d%d",&x,&y);if (dis[x][y]!=inf)  printf("%d\n",dis[x][y]);else printf("Sorry\n");}}}