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=1; k<=n; k++)
        for(i=1; i<=n; i++)
            for(j=1; 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=1; i<=n; i++)
        for(j=1; j<=m; j++)
            for(k=1; k<=m; k++)
                scanf("%d",&a[i][j][k]);
    for(i=0; i<=n; i++)
        for(j=0; j<=n; j++)
            if(i==j) map[i][j]=0;
            else   map[i][j]=INF;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
        {
            if(i==j) continue;
            memset(tem,0,sizeof(tem));
            for(x=1; x<=m; x++)
                for(y=1; y<=m; y++)
                {
                    tem[x][y]=0;
                    for(z=1; z<=m; z++)   ///矩阵计算
                        tem[x][y]+=a[i][x][z]*a[j][z][y];
                }
            for(x=1; x<=n; x++)
            {
                if(x==i||x==j)
                    continue;
                int flag=1;
                for(y=1; y<=m; y++)
                {
                    for(z=1; z<=m; z++)
                    {
                        if(tem[y][z]!=a[x][y][z]) ///比较是否完全相同
                        {
                            flag=0;
                            break;
                        }
                    }
                    if(!flag)
                        break;
                }
                if(flag)
                    map[i][x]=1;
            }
        }
    }
    floyd();
}
int main()
{
    int i,j,k,x,y;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        getmap();
        scanf("%d",&k);
        while(k--)
        {
            scanf("%d%d",&x,&y);
            if(map[x][y]<INF)
                printf("%d
",map[x][y]);
            else
                printf("Sorry
");
        }
    }
    return 0;
}