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
 
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 ;
}

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