深信服入职前编码训练21题--02

输入

5
15 7 6 6 5

输出

```mermaid
graph TD
    n0[n0:15]
    n0 --> n8
    n1[n1:7]
    n1 --> n6
    n2[n2:6]
    n2 --> n5
    n3[n3:6]
    n3 --> n6
    n4[n4:5]
    n4 --> n5
    n5((11))
    n5 --> n7
    n6((13))
    n6 --> n7
    n7((24))
    n7 --> n8
    n8((39))
```

说明
1.grath TD下面的输出都是\t开头
2.n0 ---> n8 的意思是n0的父节点是n8

  1 #include <stdio.h>
  2 #include <limits.h>
  3 #include <assert.h>
  4 #include <malloc.h>
  5 
  6 struct node {
  7     int left, right, parent;
  8     int val;
  9 };
 10 
 11 void tree_print(const struct node arr[], int cnt)
 12 {
 13     int i;
 14     for (i = 0; i < cnt; ++i) {
 15         fprintf(stderr, "%d: {left:%d,right:%d,parent:%d,val:%d}\n"
 16             , i, arr[i].left, arr[i].right, arr[i].parent, arr[i].val);
 17     }
 18 }
 19 
 20 void tree_output(FILE *fp, const struct node arr[], int old_cnt, int cnt)
 21 {
 22     int i;
 23     fprintf(fp, "```mermaid\n");
 24     fprintf(fp, "graph TD\n");
 25     for (i = 0; i < cnt; ++i) {
 26         if (i < old_cnt)
 27             fprintf(fp, "\tn%d[n%d:%d]\n", i, i, arr[i].val);
 28         else
 29             fprintf(fp, "\tn%d((%d))\n", i, arr[i].val);
 30 
 31         if (arr[i].parent >= 0) {
 32             fprintf(fp, "\tn%d --> n%d\n", i, arr[i].parent);
 33         }
 34     }
 35     fprintf(fp, "```\n");
 36 }
 37 
 38 int build_tree(struct node arr[], int cnt);
 39 
 40 static int input(int **arr, int *size)
 41 {
 42     int i;
 43     int ret;
 44 
 45     ret = fscanf(stdin, "%d\n", size);
 46     if (ret != 1)
 47         return -1;
 48     if (*size <= 0)
 49         return -1;
 50     *arr = (int *)malloc(sizeof(int) * (*size));
 51     for (i = 0; i < *size; ++i) {
 52         fscanf(stdin, "%d ", &(*arr)[i]);
 53     }
 54     return 0;
 55 }
 56 int main(int argc, char *argv[])
 57 {
 58     int *vals = NULL;
 59     int cnt = 0;
 60     struct node *arr;
 61     int i;
 62 
 63     if (input(&vals, &cnt) < 0) {
 64         fprintf(stderr, "input error\n");
 65         return 0;
 66     }
 67     arr = (struct node *)malloc(sizeof(struct node) * cnt * 3);
 68 
 69     for (i = 0; i < cnt; ++i) {
 70         arr[i].left = -1;
 71         arr[i].right = -1;
 72         arr[i].parent = -1;
 73         arr[i].val = vals[i];
 74     }
 75 
 76     int newcnt = build_tree(arr, cnt);
 77     tree_output(stdout, arr, cnt, newcnt);
 78     free(vals);
 79     free(arr);
 80     return 0;
 81 }
 82 
 83 //  建树，返回树的根节点
 84 int build_tree(struct node arr[], int cnt)
 85 {
 86     while (1) {
 87         int i;
 88         int min1 = -1;              //权值最小的节点编号
 89         int min2 = -1;              //权值第二小的节点编号
 90         int root_node = 0;          //根节点(没有父节点)的个数
 91 
 92         for (i = 0; i < cnt; ++i) {
 93             if (arr[i].parent>= 0)    //拥有父节点,则跳过这个结点
 94                 continue;
 95             ++root_node;        //根节点加1
 96             if (min1 < 0) {
 97                 min1 = i;
 98             } else if (arr[i].val < arr[min1].val) {    //出现比第一更小的,更新
 99                 min2 = min1;
100                 min1 = i;
101             } else if (min2 < 0) {
102                 min2 = i;
103             } else if (arr[i].val < arr[min2].val) {    //当前节点权值比第二小节点更小
104                 min2 = i;
105             }
106         }
107         if (root_node < 2)
108             break;
109         arr[cnt].left = min2;
110         arr[cnt].right = min1;
111         arr[cnt].val = arr[min1].val + arr[min2].val;
112         arr[cnt].parent = -1;
113         arr[min1].parent = cnt;
114         arr[min2].parent = cnt;
115         ++cnt;
116     }
117     return cnt;
118 }