poj1426 - Find The Multiple [bfs 记录路径]

转：http://blog.csdn.net/wangjian8006/article/details/7460523
(比较好的记录路径方案)
  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<string>
 12 
 13 #define N 1000005
 14 #define M 9
 15 #define mod 1000000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define LL long long
 20 #define eps 1e-6
 21 #define inf 1000000
 22 #define maxi(a,b) (a)>(b)? (a) : (b)
 23 #define mini(a,b) (a)<(b)? (a) : (b)
 24 
 25 using namespace std;
 26 
 27 int n;
 28 typedef struct
 29 {
 30     int v;
 31     int yu;
 32     int num;
 33     int pre;
 34 }PP;
 35 
 36 int tot;
 37 int vis[1005];
 38 PP ans[N];
 39 
 40 void ini()
 41 {
 42     tot=0;
 43     memset(vis,0,sizeof(vis));
 44 }
 45 
 46 void out(int now)
 47 {
 48    // for(int i=0;i<=1;i++)
 49     //printf(" v=%d pre=%d\n",ans[i].v,ans[i].pre);
 50     //printf(" now=%d v=%d pre=%d\n",now,ans[now].v,ans[now].pre);
 51     if(ans[now].pre!=-1){
 52 
 53         out(ans[now].pre);
 54 
 55     }
 56     printf("%d",ans[now].v);
 57 }
 58 
 59 void bfs()
 60 {
 61     queue<PP>q;
 62     PP te,nt;
 63     te.v=1;
 64     te.pre=-1;
 65     te.num=0;
 66     te.yu=1%n;
 67     ans[0].v=1;
 68     ans[0].pre=-1;
 69     tot=1;
 70     vis[1]=1;
 71     q.push(te);
 72 
 73     while(q.size()>=1)
 74     {
 75 
 76         te=q.front();
 77         q.pop();
 78        // printf(" v=%d yu=%d num=%d pre=%d\n",te.v,te.yu,te.num,ans[te.num].pre);
 79         if(te.yu==0){
 80             //printf(" num=%d\n",te.num);
 81             out(te.num);
 82             printf("\n");
 83             return;
 84         }
 85         nt.pre=te.num;
 86         nt.yu=(te.yu*10)%n;
 87         if(vis[nt.yu]==0){
 88             vis[nt.yu]=1;
 89             nt.num=tot;
 90             ans[tot].v=0;
 91             ans[tot].pre=te.num;
 92             tot++;
 93 
 94             nt.v=0;
 95             q.push(nt);
 96         }
 97         nt.yu=(te.yu*10+1)%n;
 98         if(vis[nt.yu]==0){
 99             vis[nt.yu]=1;
100             nt.num=tot;
101             ans[tot].v=1;
102             ans[tot].pre=te.num;
103             tot++;
104             nt.v=1;
105             q.push(nt);
106         }
107     }
108 }
109 
110 void solve()
111 {
112     bfs();
113 }
114 
115 int main()
116 {
117     //freopen("data.in","r",stdin);
118    // freopen("data.out","w",stdout);
119     //scanf("%d",&T);
120     //for(int ccnt=1;ccnt<=T;ccnt++)
121     //while(T--)
122     while(scanf("%d",&n)!=EOF)
123     {
124         if(n==0) break;
125         ini();
126         solve();
127 //        out();
128     }
129 
130     return 0;
131 }
posted on 2015-01-22 19:34 njczy2010 阅读(127) 评论(0) 编辑收藏举报