[2016-02-19][UVA][524][Prime Ring Problem]

[2016-02-19][UVA][524][Prime Ring Problem]

UVA - 524 Prime Ring Problem Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu Submit Status Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1. Input  n (0 < n <= 16) Output  The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. You are to write a program that completes above process. Sample Input  6 8 Sample Output  Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2 Miguel A. Revilla 1999-01-11

• 时间:2016-02-19 21:16:47 星期五
  
• 题目编号:UVA 524    
  
• 题目大意:给定数字n,输出1~n的数字组成的排列,使得这些排列相邻的数字之和是素数(首位相连)
  
• 方法:dfs枚举答案,回溯优化.
  
• 解题过程遇到问题:
  
• 格式问题:最后一组数据后面没有空行,否则WA
• 每行最后一个数字后面没有空格,否则PE  

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> using namespace std; typedef long long LL; #define CLR(x,y) memset((x),(y),sizeof((x))) #define FOR(x,y,z) for(int (x)=(y);(x)<(z);(x)++) #define FORD(x,y,z) for(int (x)=(y);(x)>=(z);(x)--) #define FOR2(x,y,z) for((x)=(y);(x)<(z);(x)++) #define FORD2(x,y,z) for((x)=(y);(x)>=(z);(x)--) const int maxn = 20; int ispri[maxn*2],n,vis[maxn],res[maxn]; void setpri(){         ispri[1] = 1;         CLR(ispri,-1);         FOR(i,2,maxn*2){                 if(ispri[i]){                         for(int j = i + i;j < maxn*2;j += i)                                 ispri[j] = 0;                 }         } } void dfs(int cur){         if(cur == n ){                 if(!ispri[res[0] + res[n - 1]]) return ;                 printf("%d",res[0]);                 FOR(i,1,n){                         printf(" %d",res[i]);                 }                 puts("");                 return ;         }         FOR(i,2,n+1){                 if(!vis[i] && ispri[ res[cur - 1] + i ]){                         vis[i] = 1;                         res[cur] = i;                         dfs(cur + 1);                         vis[i] = 0;                 }         } } int main(){         int cntcase = 0;         setpri();         while(~scanf("%d",&n)){                 if(cntcase)     puts("");                 printf("Case %d:\n",++cntcase);                 CLR(vis,0);                 vis[1] = res[0] = 1;                 dfs(1);         }     return 0; }

来自为知笔记(Wiz)