hdu-----(2807)The Shortest Path(矩阵+Floyd)
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): 2440 Accepted Submission(s): 784
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
题目很清楚：
如果满足A*B=C 那么就说A到C是联通的....
反之则为不连通...
这里需要用到的求最短路径，对于稠密图的，用弗洛伊德算法比狄斯喹诺算法要好一点。
至于要优化，看来结题报告之后，觉得还是不大靠谱，就没有写，写的是一个朴素的矩阵相乘算法+floyd算法
代码：
1 #include<cstdio>
2 #include<cstring>
3 #define inf 0x3f3f3f3f
4 using namespace std;
5
6 const int maxn=82;
7 int arr[maxn][maxn][maxn];
8 int ans[maxn][maxn];
9 int tem[maxn][maxn];
10
11 int n,m,w;
12
13 void init(int a[][maxn])
14 {
15 for(int i=1;i<=n;i++) //城市初始化
16 {
17 for(int j=1;j<=n;j++)
18 {
19 if(i==j)a[i][j]=0;
20 else a[i][j]=inf;
21 }
22 }
23 }
24
25 void floyd(int a[][maxn]) //运用floyd算法求城市间的最短路径
26 {
27 for(int k=1;k<=n;k++)
28 {
29 for(int i=1;i<=n;i++)
30 {
31 for(int j=1;j<=n;j++)
32 {
33 if(ans[i][j]>ans[i][k]+ans[k][j])
34 ans[i][j]=ans[i][k]+ans[k][j];
35 }
36 }
37 }
38 }
39
40 void Matrix(int a[][maxn],int p1,int p2)
41 {
42 for(int i=1;i<=m;i++)
43 {
44 for(int j=1;j<=m;j++)
45 {
46 a[i][j]=0; // init()
47 for(int k=1;k<=m;k++)
48 {
49 a[i][j]+=arr[p1][i][k]*arr[p2][k][j];
50 }
51 }
52 }
53 }
54
55 void work()
56 {
57 int t1,t2,t3;
58 for(int i=1;i<=n;i++)
59 {
60 for(int j=1;j<=n;j++)
61 {
62 if(i==j) continue; //a，b 两数组不能相同
63 Matrix(tem,i,j); //两个矩阵相乘
64 for(t1=1;t1<=n;t1++)
65 {
66 //a，b,c三数组不能相同
67 if(t1!=i&&t1!=j)
68 {
69 for( t2=1;t2<=m;t2++)
70 {
71 for(t3=1;t3<=m;t3++)
72 {
73 //得到的结果相比较
74 if(tem[t2][t3]!=arr[t1][t2][t3])
75 goto loop;
76 }
77 }
78 loop:
79 if(t3>m)
80 ans[i][t1]=1;
81 }
82 }
83 }
84 }
85 }
86
87 int main()
88 {
89 int a,b;
90 while(scanf("%d%d",&n,&m)&&n+m!=0)
91 {
92 for(int i=1;i<=n;i++)
93 for(int j=1;j<=m;j++)
94 for(int k=1;k<=m;k++)
95 scanf("%d",&arr[i][j][k]);
96 init(ans);
97 work();
98 floyd(ans);
99 scanf("%d",&w);
100 while(w--)
101 {
102 scanf("%d%d",&a,&b);
103 if(ans[a][b]==inf)
104 printf("Sorry\n");
105 else
106 printf("%d\n",ans[a][b]);
107 }
108 }
109 return 0;
110 }