POJ 1351 Number of Locks (记忆化搜索 状态压缩)

Number of Locks

Description

In certain factory a kind of spring locks is manufactured. There are n slots (1 < n < 17, n is a natural number.) for each lock. The height of each slot may be any one of the 4 values in｛1,2,3,4｝( neglect unit ). Among the slots
of a lock there are at least one pair of neighboring slots with their difference of height equal to 3 and also there are at least 3 different height values of the slots for a lock. If a batch of locks is manufactured by taking all over the 4 values for slot
height and meet the two limitations above, find the number of the locks produced.

Input

There is one given data n (number of slots) on every line. At the end of all the input data is -1, which means the end of input.

Output

According to the input data, count the number of locks. Each output occupies one line. Its fore part is a repetition of the input data and then followed by a colon and a space. The last part of it is the number of the locks counted.

Sample Input

2
3
-1

Sample Output

2: 0
3: 8

Source

Xi‘an 2002

题目链接：http://poj.org/problem?id=1351

题目大意：用数字1，2，3，4求一个长度为n的序列，要求这个序列中至少有三个不同的数字且至少有一组相邻的值相差3，求满足条件的序列的个数

题目分析：dp[num][t][ok][last]表示当前序列长度为num，用了t种数字，若有有相邻的1，4则ok为1，否则为0，last表示当前序列的最后一个数，采用记忆化搜索，DFS里除了dp的4个值还要多一个参数st，用2进制表示当前数字的使用状态，例如st＝1111表示四个数都用了，每次将搜索的数与当前状态与一下来判断使用个数有没有增加.

#include <cstdio>
#include <cstring>
#define ll long long
ll dp[20][5][2][5];
int n;

ll DFS(int num, int t, int ok, int last, int st)
{
    if(num == n)
    {
        if(t >= 3 && ok)
            return 1;
        else
            return 0;
    }
    if(dp[num][t][ok][last] != -1)
        return dp[num][t][ok][last];
    ll tmp = 0;
    for(int i = 1; i <= 4; i++)
    {
        int tt, ok2 = 0, curst = 1 << (i - 1);
        if((i == 1 && last == 4) || (i == 4 && last == 1))
            ok2 = 1;
        if(st & curst)
            tt = t;
        else
            tt = t + 1;
        if(tt > 3)
            tt = 3;
        tmp += DFS(num + 1, tt, ok || ok2, i, st | curst);
    }
    return dp[num][t][ok][last] = tmp;
}

int main()
{
    while(scanf("%d", &n) != EOF && n != -1)
    {
        memset(dp, -1, sizeof(dp));
        printf("%d: %lld\n", n, DFS(0, 0, 0, 0, 0));
    }
}