hdu Sequence one

Sequence one

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 158    Accepted Submission(s): 49

Problem Description

Search is important in the acm algorithm. When you want to solve a problem by using the search method, try to cut is very important. Now give you a number sequence, include n (<=1000) integers, each integer not bigger than 2^31, you want to find the first P subsequences that is not decrease (if total subsequence W is smaller than P, than just give the first W subsequences). The order of subsequences is that: first order the length of the subsequence. Second order the sequence of each integer’s position in the initial sequence. For example initial sequence 1 3 2 the total legal subsequences is 5. According to order is {1}; {3}; {2}; {1,3}; {1,2}. {1,3} is first than {1,2} because the sequence of each integer’s position in the initial sequence are {1,2} and {1,3}. {1,2} is smaller than {1,3}. If you also can not understand , please see the sample carefully.

 

Input

The input contains multiple test cases. Each test case include, first two integers n, P. (1<n<=1000, 1<p<=10000).

 

Output

For each test case output the sequences according to the problem description. And at the end of each case follow a empty line.

 

Sample Input

3 5
1 3 2
3 6
1 3 2
4 100
1 2 3 2

 

Sample Output

1
3
2
1 3
1 2

1
3
2
1 3
1 2

1
2
3
1 2
1 3
2 3
2 2
1 2 3
1 2 2

Hint

Hint : You must make sure each subsequence in the subsequences is unique.

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

分析：通过枚举子串长度分别dfs知道没有输出或已达到要求个数，并且保证每次进入dfs采用的数都不同。

#include<cstdio>
#include<cstring>
int cnt;
int a[1010];
int ans[1010];
int tot;
int n, k;
int flag;

int fun(int x, int y) {
    int i;
    for (i = x; i < y; ++i)
        if (a[i] == a[y])
            return 0;
    return 1;
}

void dfs(int now, int num) {
    int i;
    if (num == tot) {
        ++cnt;
        for (i = 0; i < tot - 1; ++i)
            printf("%d ", ans[i]);
        printf("%d\n", ans[tot - 1]);  
        if (cnt == k) 
        flag = 1;
        return;
    }

    if (num == 0) {
   //     printf("**");
        for (i = 0; i < n; ++i) {
          //  printf("%d ",i);
            if (fun(0, i)) {  
           //    printf("%d ",i); 
               ans[num]=a[i];
                dfs(i, 1);
                if (flag)
                    return;
            }
        }
    } else {
        for (i = now + 1; i < n; ++i) {
            if (a[i] >= a[now] && fun(now + 1, i)) {
                ans[num]=a[i];
                dfs(i, num + 1);
                if (flag)
                    return;
            }
        }
    }
}

int main() {
    int i, t;
    while (scanf("%d%d", &n, &k) != EOF) {
        for (i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        cnt = 0;
        for (tot = 1; cnt < k; ++tot) {
          //  printf("tot=%d\n",tot);
            flag = 0;
            t = cnt;
            dfs(0, 0);
            if(t==cnt)
                break;
            if (flag)
                break;
        }
        printf("\n");
    }
    return 0;
}