PAT 1004. Counting Leaves (30)

链接： http://www.patest.cn/contests/pat-a-practise/1004

复习一下C

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<ctype.h>
  4 #include<stdlib.h>
  5 
  6 #define MAX_NODE 110
  7 #define MAX_QUEUE_SIZE 110
  8 
  9 typedef struct {
 10     int *child;
 11     int child_size;
 12     int deep;
 13 } tree_node;
 14 tree_node nodes[MAX_NODE];
 15 
 16 int result_cnt;
 17 int result[MAX_NODE];
 18 
 19 int rear = -1;
 20 int front = -1;
 21 int queue[MAX_QUEUE_SIZE];
 22 
 23 void addq(int *rear, int item)
 24 {
 25     if (*rear == MAX_QUEUE_SIZE - 1) { //full queue
 26         return;
 27     }
 28     queue[++*rear] = item;
 29 }
 30 
 31 int deleteq(int *front, const int rear)
 32 {
 33     if (*front == rear) { //empty queue
 34         return;
 35     }
 36     return queue[++*front];
 37 }
 38 
 39 int is_empty()
 40 {
 41     return front == rear;
 42 }
 43 
 44 void initq()
 45 {
 46     memset(queue, 0, sizeof(queue));
 47     rear = front = -1;
 48 }
 49 
 50 void level_order()
 51 {
 52     int i;
 53     initq();
 54     addq(&rear, 1);
 55     nodes[1].deep = 0;
 56     while(!is_empty()) {
 57         int u = deleteq(&front, rear);
 58         int deep = nodes[u].deep;
 59         if (0 == nodes[u].child_size) {
 60             ++result[deep];
 61             result_cnt = deep;
 62             continue;
 63         }
 64         for (i = 0; i < nodes[u].child_size; ++i) {
 65             addq(&rear, nodes[u].child[i]);
 66             nodes[nodes[u].child[i]].deep = deep + 1;
 67         }
 68     }
 69 }
 70 
 71 void init()
 72 {
 73     int i = 0;
 74     memset(result, 0, sizeof(result));
 75     result_cnt = 0;
 76     for (i = 1; i < MAX_NODE; ++i) {
 77         nodes[i].child = NULL;
 78         nodes[i].child_size = 0;
 79         nodes[i].deep = 0;
 80     }
 81 }
 82 int main(int argc, char **argv)
 83 {
 84     init();
 85     int n, m, i, j;
 86     int id, k, child_id;
 87     scanf("%d%d", &n, &m);
 88     for (i = 0; i < m; ++i) {
 89         scanf("%d%d", &id, &k);
 90         nodes[id].child_size = k;
 91         nodes[id].child = (int *)malloc(sizeof(int)*k);
 92         for (j = 0; j < k; ++j) {
 93             scanf("%d", &child_id);
 94             nodes[id].child[j] = child_id;
 95         }
 96     }
 97     level_order();
 98     printf("%d", result[0]);
 99     for (i = 1; i <= result_cnt; ++i) {
100         printf(" %d", result[i]);
101     }
102     puts("");
103     return 0;
104 }

 

posted @ 2015-07-25 04:31  ACSeed  Views(137)  Comments(0)    收藏  举报