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UVA - 524 Prime Ring Problem(素数环)(回溯法)

题意:输入n,把1~n组成个环,相邻两个数之和为素数。

分析:回溯法。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int prime[MAXN];
int a[MAXN];
int vis[MAXN];
int n;
void init(){
    for(int i = 2; i < MAXN; ++i) prime[i] = 1;
    for(int i = 2; i < MAXN; ++i){
        if(prime[i]){
            for(int j = i + i; j < MAXN; j += i){
                prime[j] = 0;
            }
        }
    }
}
void dfs(int cur){
    if(cur == n && prime[a[n] + a[1]]){
        for(int i = 1; i <= n; ++i){
            if(i != 1) printf(" ");
            printf("%d", a[i]);
        }
        printf("\n");
    }
    else{
        for(int i = 2; i <= n; ++i){
            if(!vis[i] && prime[a[cur] + i]){
                vis[i] = 1;
                a[cur + 1] = i;
                dfs(cur + 1);
                vis[i] = 0;
            }
        }
    }
}
int main(){
    init();
    a[1] = 1;
    int kase = 0;
    while(scanf("%d", &n) == 1){
         if(kase) printf("\n");
         printf("Case %d:\n", ++kase);
         dfs(1);
    }
    return 0;
}

 

UVA - 524 Prime Ring Problem(素数环)(回溯法)