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  • hdu 3999 The order of a Tree

    先建立一颗二叉搜索树,然后先序遍历一下,就可以出答案了。

    我用了比较奇葩的写法。。。二叉搜索树结构体模拟了一下。先序遍历用的DFS。。。。。。

    #include<stdio.h>
    #include<string.h>
    struct abc{
        int left;
        int right;
        int date;
    }node[100010];
    int flag[100010];
    int ans[100010];
    int n, tot;
    void dfs(int wei)
    {
        if (node[wei].left != -1 && flag[node[wei].left] ==0)
        {
            flag[node[wei].left] = 1;
            ans[tot] = node[node[wei].left].date;
            tot++;
            dfs(node[wei].left);
        }
        if (node[wei].right != -1 && flag[node[wei].right] == 0)
        {
            flag[node[wei].right] = 1;
            ans[tot] = node[node[wei].right].date;
            tot++;
            dfs(node[wei].right);
        }
    }
    int main()
    {
        int i,  k;
        while (~scanf("%d", &n))
        {
            memset(flag, 0, sizeof(flag));
            for (i = 0; i <= n; i++) node[i].date = -1, node[i].left = -1, node[i].right = -1;
            scanf("%d", &k);
            tot = 1;
            node[1].date = k;
            int nodesum = 2;
            for (i = 2; i <= n; i++)
            {
                scanf("%d", &k);
                int nodexun = 1;
                while (1)
                {
                    if (k > node[nodexun].date)
                    {
                        if (node[nodexun].right == -1)
                        {
                            node[nodexun].right = nodesum;
                            node[nodesum].date = k;
                            nodesum++;
                            break;
                        }
                        else nodexun = node[nodexun].right;
                    }
                    else
                    {
                        if (node[nodexun].left == -1)
                        {
                            node[nodexun].left = nodesum;
                            node[nodesum].date = k;
                            nodesum++;
                            break;
                        }
                        else nodexun = node[nodexun].left;
                    }
                }
            }
            ans[tot] = node[1].date;
            tot++;
            flag[1] = 1;
            dfs(1);
            for (i = 1; i <= n; i++)
            {
                if (i<n)
                printf("%d ", ans[i]);
                else printf("%d
    ", ans[i]);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/4438834.html
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