The Shortest Path
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2623 Accepted Submission(s): 827
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?
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
Recommend
这道题是比较单纯的一道题，暴力可解。不过写不好容易超时。
题意是每个城市用一个矩阵来表示，如果A*B=C,这说明A和C之间存在一条有向边。然后建图，用floyd算法求出所有点之间的最短路径。给定两个点x,y,判断这两点之间是否存在道路，如果存在输出这两点之间的最短路径。判断两点之间是否存在道路，这个属于图的传递闭包，是用floyd算法求出的。但是我们用floyd算法求解最短路径时，如果算出x和y之间的距离为INF(无穷大)，那么也就可以说明两点之间不存在道路，不可到达。也就不需要求传递闭包。
在暴力求解的过程中需要注意的是在进行A*B=C的判断时要保证A,B,C互不相同。而且计算出A*B以后，要逐个比较，看其是否与A*B所得矩阵相等。不能一旦找到以后就break，因为有可能有多个矩阵都等于A*B。这几点要注意。
参考代码：
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<ctime>
#include<cstdlib>
#include<iomanip>
#include<utility>
#define pb push_back
#define mp make_pair
#define CLR(x) memset(x,0,sizeof(x))
#define _CLR(x) memset(x,-1,sizeof(x))
#define REP(i,n) for(int i=0;i<n;i++)
#define Debug(x) cout<<#x<<"="<<x<<" "<<endl
#define REP(i,l,r) for(int i=l;i<=r;i++)
#define rep(i,l,r) for(int i=l;i<r;i++)
#define RREP(i,l,r) for(int i=l;i>=r;i--)
#define rrep(i,l,r) for(int i=1;i>r;i--)
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<11
using namespace std;
const int INF=(1<<30);
int n,m,t;
int g[100][100];
struct mat
{
int d[100][100];
} A[100];
int main()
{
while(~scanf("%d%d",&n,&m)&&n+m)
{
REP(i,1,n)
rep(j,0,m)
rep(k,0,m)
scanf("%d",&A[i].d[j][k]);
REP(i,1,n) //初始化邻接矩阵，对于同一个点时，距离为0(注意本题没有自环的情形，因为A,B,C互不相等，所以同一个点到同一个点距离初始化为0)
{
REP(j,i,n)
{
if(i==j) g[i][j]=0;
else g[i][j]=g[j][i]=INF;
}
}
REP(i,1,n)
{
REP(j,1,n)
{
if(i!=j)
{
mat m1;
rep(l1,0,m) //矩阵乘法求解A*B
{
rep(l2,0,m)
{
m1.d[l1][l2]=0;
rep(l3,0,m)
{
m1.d[l1][l2]+=A[i].d[l1][l3]*A[j].d[l3][l2];
}
}
}
REP(k,1,n) //进行矩阵的比较
{
if(i!=k&&j!=k)
{
bool flag=1;
rep(l1,0,m)
{
rep(l2,0,m)
{
if(m1.d[l1][l2]!=A[k].d[l1][l2])
{
flag=0;
break;
}
}
if(!flag)
break;
}
if(flag)
g[i][k]=1;
}
}
}
}
}
REP(k,1,n) //floyd算法求最短路径
REP(i,1,n)
{
if(g[i][k]==INF)
continue;
REP(j,1,n)
if(g[i][k]<INF&&g[k][j]<INF)
g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
}
scanf("%d",&t);
while(t--)
{
int u,v;
scanf("%d%d",&u,&v);
if(g[u][v]<INF)
printf("%d\n",g[u][v]);
else
printf("Sorry\n");
}
}
}
