SDUT 2139 图结构练习——BFS——从起始点到目标点的最短步数

 题目链接：http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2139

  裸BFS。

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 int p[1001][1001],o[1001],k[1001];
 4 int bfs(int n)
 5 {
 6     int i,j,start,end,a,num = 1;
 7     start = end = 0;
 8     while(start <= end)
 9     {
10        a = 1;
11        for(i = start;i <= end;i ++)
12        {
13           for(j = 1;j <= n;j ++)
14           {
15               if(p[k[i]][j] == 1 && o[j]!=1)
16               {
17                   k[end+a] = j;
18                   o[j] = 1;
19                   a ++;
20               }
21           }
22        }
23        start = end + 1;
24        end =end + a -1;
25        for(i = start;i <= end;i ++)
26        {
27            if(k[i] == 1)
28            {
29                return num;
30            }
31        }
32        num ++;
33     }
34     return 0;
35 }
36 int main()
37 {
38     int i,n,m,a,b;
39     while(scanf("%d%d",&n,&m)!=EOF)
40     {
41         memset(k,0,sizeof(k));
42         memset(p,0,sizeof(p));
43         memset(o,0,sizeof(o));
44         for(i = 1;i <= m;i ++)
45         {
46             scanf("%d%d",&a,&b);
47             p[a][b] = 1;
48         }
49         k[0] = n;o[n] = 1;
50         i = bfs(n);
51         if(i == 0)
52         printf("NO\n");
53         else
54         printf("%d\n",i);
55     }
56     return 0;
57 }

 

posted @ 2012-05-29 15:18  Naix_x  阅读(213)  评论(0)  编辑  收藏  举报