2017 ICPC/ACM 沈阳区域赛HDU6223

Posted 2020-10-14 灬从此以后灬

Infinite Fraction Path

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1262    Accepted Submission(s): 224

Problem Description

The ant Welly now dedicates himself to urban infrastructure. He came to the kingdom of numbers and solicited an audience with the king. He recounted how he had built a happy path in the kingdom of happiness. The king affirmed Welly’s talent and hoped that this talent can help him find the best infinite fraction path before the anniversary.
The kingdom has N cities numbered from 0 to N - 1 and you are given an array D[0 ... N - 1] of decimal digits (0 ≤ D[i] ≤ 9, D[i] is an integer). The destination of the only one-way road start from the i-th city is the city labelled (i2 + 1)%N.
A path beginning from the i-th city would pass through the cities u1,u2,u3, and so on consecutively. The path constructs a real number A[i], called the relevant fraction such that the integer part of it is equal to zero and its fractional part is an infinite decimal fraction with digits D[i], D[u1], D[u2], and so on.
The best infinite fraction path is the one with the largest relevant fraction

 

 

Input

The input contains multiple test cases and the first line provides an integer up to 100 indicating to the total numberof test cases.
For each test case, the first line contains the integer N (1 ≤ N ≤ 150000). The second line contains an array ofdigits D, given without spaces.
The summation of N is smaller than 2000000.

 

 

Output

For each test case, you should output the label of the case first. Then you are to output exactly N characters which are the first N digits of the fractional part of the largest relevant fraction.

 

 

Sample Input

4

3

149

5

12345

7

3214567

9

261025520

 

Sample Output

Case #1: 999

Case #2: 53123

Case #3: 7166666

Case #4: 615015015

 

分析   这道题写不出来 搜到大佬的题解 只有代码分析很少  只能看懂大佬代码  然后照着 敲一遍 不得不服大佬就是大佬写的代码很有感觉

 

按照 i ——> ( i ^ 2 + 1 ) % n  建图    bfs+剪枝

bfs + 剪枝 

剪枝：

• 值小于当前层最大值的点移出队列
• 同一层在相同位置的移出队列

 

大牛博客   http://blog.csdn.net/qq_40482495/article/details/78492841

AC代码

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <vector>
11 using namespace std;
12 const int maxn= 150000+10;
13 const int maxm= 1e4+10;
14 const int inf = 0x3f3f3f3f;
15 typedef long long ll;
16 ll ne[maxn];   //ne[i]为以i为起点的边的终点
17 int v[maxn],ans[maxn],vis[maxn];  //权值、答案、i上一次在ans数组中出现的位置
18 char s[maxn];
19 int n,t;
20 struct node
21 {
22     int v,pos,ans;   //表示当前节点的权值，下标i,在答案数组中的位置下标
23     node(){}
24     node(int v,int pos,int ans):v(v),pos(pos),ans(ans){}
25 };
26 struct compare
27 {
28     bool operator()(const node &a,const node &b) const   //ans最小值优先 权值最大值优先 
29     {
30         if(a.ans!=b.ans) return a.ans>b.ans;       
31         else if(a.v!=b.v) return a.v<b.v;
32         return a.pos>b.pos;
33     }
34 };
35 priority_queue<node,vector<node>,compare> q;
36 int main()
37 {
38     scanf("%d",&t);
39     int kase=1;
40     while(t--)
41     {
42         scanf("%d",&n);
43         scanf("%s",s);
44         memset(vis,-1,sizeof(vis));    //初始化
45         memset(ans,-1,sizeof(ans));
46         int ma=0;
47         for(int i=0;i<n;i++)            //权值转化为整数  求出最大值  i的终点(i^2+1)%n
48         { 
49             v[i]=s[i]-‘0‘;
50             ma=max(ma,v[i]);      
51             ne[i]=(((ll)i*(ll)i+1)%(ll)n);  
52         }
53 //        for(int i=0;i<n;i++)
54 //        {
55 //            printf("%d %d %d\n",i,v[i],next[i]);
56 //        }
57 //        printf("%d\n",ma);
58         for(int i=0;i<n;i++)
59         {
60             if(v[i]==ma)
61                 q.push(node(ma,i,0));       //最大值先压入队列
62         }
63         while(!q.empty())
64         {
65             node t=q.top();q.pop();
66             if(ans[t.ans]==-1) ans[t.ans]=t.v;    //该位置初步确定一个值
67             if(ans[t.ans]>t.v) continue;          //该节点的权值比以前小  直接跳过
68             if(vis[t.pos]<t.ans) vis[t.pos]=t.ans;   //更新节点的 访问位置
69             else continue;
70             if(t.ans==n-1) continue;
71             q.push(node(v[ne[t.pos]],ne[t.pos],t.ans+1));  //加入新节点
72         }
73         printf("Case #%d: ",kase++);
74         for(int i=0;i<n;i++)
75             printf("%d",ans[i]);
76         printf("\n");
77     }
78     return 0;
79 }