Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D

The Grand Dinner

Input: standard input

Output: standard output

Time Limit: 15 seconds

Memory Limit: 32 MB

 

Each team participating in this year’s ACM World Finals contest is expected to join the grand dinner to be arranged after the prize giving ceremony ends. In order to maximize the interaction among the members of different teams, it is expected that no two members of the same team sit at the same table.

 

Now, given the number of members in each team (including contestants, coaches, reserves, guests etc.) and the seating capacity of each available table, you are to determine whether it is possible for the teams to sit as described in the previous paragraph. If such an arrangement is possible you must also output one possible seating arrangement. If there are multiple possible arrangements, any one is acceptable.

 

Input

 The input file may contain multiple test cases. The first line of each test case contains two integers M (1 £ M £ 70) and N (1 £ N £ 50) denoting the number of teams and the number of tables respectively. The second line of the test case contains M integers where the i-th (1 £ i £ M) integer m_i (1 £ m_i £ 100) indicates the number of members of team i. The third line contains N integers where the j-th (1 £ j £ N) integer n_j (2 £ n_j £ 100) indicates the seating capacity of table j.

 

A test case containing two zeros for M and N terminates the input.

 

Output

For each test case in the input print a line containing either 1 or 0 depending on whether or not there exists a valid seating arrangement of the team members. In case of a successful arrangement print M additional lines where the i-th (1 £ i £ M) of these lines contains a table number (an integer from 1 to N) for each of the members of team i.

 

Sample Input

4 5

4 5 3 5

3 5 2 6 4

4 5

4 5 3 5

3 5 2 6 3

0 0

 

 

Sample Output

1

1 2 4 5

1 2 3 4 5

2 4 5

1 2 3 4 5

0

(World Finals Warm-up Contest, Problem Setter: Rezaul Alam Chowdhury)

每次抓最多人的團體出來分桌, 分桌的時候以剩餘最多的桌子開始分配

#include <stdio.h>
#include <stdlib.h>
typedef struct {
    int c, i;
} DD;
int cmp(const void *i, const void *j) {
    DD *x, *y;
    x = (DD *)i, y = (DD *)j;
    if(x->c != y->c)
        return y->c - x->c;
    return x->i - y->i;
}
int cmp1(const void *i, const void *j) {
    return *(int *)i - *(int *)j;
}
int main() {
    int n, m, i, j, B[70];
    DD A[70];
    DD D[50];
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0 && m == 0)    break;
        for(i = 0; i < n; i++)
            scanf("%d", &A[i].c), A[i].i = i, B[i] = A[i].c;
        for(i = 0; i < m; i++)
            scanf("%d", &D[i].c), D[i].i = i;
        int ans[70][70];
        qsort(A, n, sizeof(DD), cmp);
        for(i = 0; i < n; i++) {
            qsort(D, m, sizeof(DD), cmp);
            int tmp = A[i].c, idx = 0, err = 0;
            for(j = 0; j < m; j++) {
                if(tmp == 0)    break;
                if(D[j].c == 0) {
                    err = 1;
                    break;
                } else {
                    D[j].c--;
                    tmp--;
                    ans[A[i].i][idx++] = D[j].i;
                }
            }
            if(err || tmp) break;
        }
        if(i == n) {
            puts("1");
            for(i = 0; i < n; i++) {
                qsort(ans[i], B[i], sizeof(int), cmp1);
                printf("%d", ans[i][0]+1);
                for(j = 1; j < B[i]; j++)
                    printf(" %d", ans[i][j]+1);
                puts("");
            }
        } else
            puts("0");
    }
    return 0;
}