UVa 524 Prime Ring Problem(素数环)
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题目链接:UVa 524
题意:
输入正整数n,把正整数1~n组成一个环,使得相邻两个整数之和均为素数。输出时从整数1开始逆时针排列。同一个环应恰好输出一次。n<=16.
分析:
简单BFS.
注意:
- 两测试案例之间有一行空格,输出数据之间有空格,行末无空格
CODE:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <algorithm>#include <queue>using namespace std;int prime[40] = { 0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1 };int n, cases = 0, first = 1;struct Node {int a, cnt;int vis[20];int num[20];}cur,nextnode;void bfs(){queue<Node> q;memset(cur.vis, 0, sizeof(cur.vis));memset(cur.num, 0, sizeof(cur.num));cur.vis[1] = cur.a = cur.cnt = 1;cur.num[1] = 1;q.push(cur);while (!q.empty()){cur = q.front();q.pop();for (int i = 2; i <= n; i++){if (cur.vis[i] || !prime[cur.a + i]) continue;nextnode.a = i;nextnode.cnt = cur.cnt + 1;for (int j = 1; j <= n; j++){nextnode.vis[j] = cur.vis[j];nextnode.num[j] = cur.num[j];}nextnode.vis[i] = 1;nextnode.num[nextnode.cnt] = i;if (nextnode.cnt == n && prime[nextnode.num[n] + 1]){cout << 1;for (int j = 2; j <= n; j++)cout << " "<< nextnode.num[j] ;putchar('\n');}q.push(nextnode);}}}int main(){#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);#endifwhile (~scanf("%d",&n)){if (first) first = 0;else printf("\n");printf("Case %d:\n", ++cases);bfs();}return 0;}
DFS写法:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <memory>using namespace std;int prime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0};int vis[20],n,cases=0,a[20];//a[i]=j:素数环第i个位置是j(下标从1开始)void dfs(int x)//处理第x个位置{ if(x-1==n) { if(prime[a[n]+1]) { for(int i=1;i<=n;i++) { if(i==1) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); } return ; } else for(int i=2;i<=n;i++) { if(!vis[i]&&prime[i+a[x-1]]) { vis[i]=1; a[x]=i; dfs(x+1); vis[i]=0; } }}int main(){#ifdef LOCAL //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout);#endif int first=1; while(cin>>n) { if(first) first=0; else cout<<endl; printf("Case %d:\n",++cases); a[1]=1; memset(vis,0,sizeof(vis)); dfs(2); } return 0;}
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