http://poj.org/problem?id=2034

大致题意：给出区间[n,m],对这个区间的数进行排列使得相邻的2个、3个......d个数之和都不是素数。输出字典序最小的。

思路：裸的dfs。TLE了无数次是因为素数打表的范围太小，最大应打到10000。

#include <stdio.h>
#include <iostream>
#include <map>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-8
#define PI acos(-1.0)
using namespace std;

const int maxn = 10010;

bool prime[maxn];
int vis[1010],ans[1010];
int n,m,d;
int ok;

void init()
{
    memset(prime,true,sizeof(prime));
    prime[0] = prime[1] = false;

    for(int i = 2; i <= 10000; i++)
    {
        if(prime[i] == true)
        {
            for(int j = i*i; j <= 10000; j += i)
                prime[j] = false;
        }
    }
}

void dfs(int dep)
{
	if(ok == 1)
		return;
	if(dep == m-n+2)
	{
		ok = 1;
		for(int i = 1; i < dep-1; i++)
			printf("%d,",ans[i]);
		printf("%d\n",ans[dep-1]);
		return;
	}

	for(int i = n; i <= m; i++)
	{
		if(!vis[i])
		{
			bool flag = 0;

			for(int k = 2; k <= d && dep-k>=0; k++)
			{
				int sum = 0;
				for(int j = dep-k+1; j <= dep-1; j++)
					sum += ans[j];
				if(prime[sum + i] == true)
					flag = 1;
			}

			if(flag == 1)
				continue;

			ans[dep] = i;
			vis[i] = 1;

			dfs(dep+1);

			vis[i] = 0;
		}
	}
}

int main()
{
    init();
    while(~scanf("%d %d %d",&n,&m,&d))
    {
        if(n == 0 && m == 0 && d == 0) break;
        memset(vis,0,sizeof(vis));
        ok = 0;

        dfs(1);

        if(ok == 0)
			printf("No anti-prime sequence exists.\n");
    }
    return 0;

}

poj 2034 Anti-prime Sequences(dfs)