uva 269 - Counting Patterns(构造)

题目链接：uva 269
- Counting Patterns

题目大意：给出n和k，要求找出满足的序列，要求为n元组，由-k到k组成，并且和为0。求出所有满足的元组个数，并且对于左移，右移，水平翻转，每个元素取相反数相同的视为一种，用字典序最大的表示，输出按照字典序最小的输出。

解题思路：因为表示的时候按照字典序最大的表示，一开始枚举开头的位置，那么在后面的数的绝对值就不会大于该数。最后判断一下，如果该序列不是最优的表示方法，就不是该情况。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;
const int maxn = 100005;
const int maxm = 15;

int s[maxn][maxm];
int n, k, key, cnt, a[maxm];

void put (int num[]) {
    printf("(%d", num[0]);
    for (int i = 1; i < n; i++)
        printf(",%d", num[i]);
    printf(")\n");
}

bool cmp (int l[], int r[]) {
    for (int i = 0; i < n; i++)
        if (l[i] != r[i])
            return l[i] > r[i];
    return false;
}

void check () {
    int b[maxm*2];

    for (int i = 0; i < n; i++)
        b[i] = b[i+n] = a[i];
    for (int i = 0; i < n; i++)
        if (b[i] == a[0] && cmp(b+i, a))
            return;

    for (int i = 0; i < 2 * n; i++)
        b[i] = -b[i];
    for (int i = 0; i < n; i++)
        if (b[i] == a[0] && cmp(b+i, a))
            return;

    for (int i = 0; i < n; i++)
        swap(b[i], b[2*n-i-1]);
    for (int i = 0;  i < n; i++)
        if (b[i] == a[0] && cmp(b+i, a))
            return;

    for (int i = 0; i < 2 * n; i++)
        b[i] = -b[i];
    for (int i = 0; i < n; i++)
        if (b[i] == a[0] && cmp(b+i, a))
            return;

    memcpy(s[cnt++], a, sizeof(a));
}

void dfs (int d, int sum) {
    if (d == n) {
        if (sum == 0)
            check();
        return;
    }

    if (abs(sum) > ((n - d) * a[0]))
        return;

    for (a[d] = -key; a[d] < key; a[d]++) {
        if (a[d-1] == key && a[d] > a[1])
            return;
        dfs(d + 1, sum + a[d]);
    }

    if (a[d] == key && a[d-1] <= a[1])
        dfs(d+1, sum+a[d]);
}

int main () {
    int cas = 0;
    while (scanf("%d%d", &n, &k) == 2 && n) {
        cnt = 1;
        for (a[0] = 1; a[0] <= k; a[0]++) {
            key = a[0];
            dfs(1, key);
        }

        if (cas++)
            printf("\n");
        printf("%d\n", cnt);
        for (int i = 0; i < cnt; i++)
            put(s[i]);
    }
    return 0;
}

uva 269 - Counting Patterns(构造)