A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.<\/p>\n
Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:<\/p>\n
ID K ID[1] ID[2] … ID[K]
\nwhere ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.<\/p>\n
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.<\/p>\n
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.<\/p>\n
2 1\n01 1 02\n<\/code><\/pre>\nSample Output:<\/h3>\n0 1\n<\/code><\/pre>\n\u9898\u76ee\u5927\u610f<\/h3>\n
\u6709\u4e00\u4e2a\u6811\uff0c\u6839\u7ed3\u70b9\u7f16\u53f7\u4e3a01\uff0c\u4f60\u9700\u8981\u627e\u51fa\u6bcf\u4e00\u5c42\u7684\u53f6\u5b50\u7ed3\u70b9\u7684\u4e2a\u6570\u3002N\u662f\u7ed3\u70b9\u7684\u4e2a\u6570\uff0cM\u662f\u975e\u53f6\u5b50\u7ed3\u70b9\u7684\u4e2a\u6570\u3002\u7b2c\u4e00\u884c\u6570\u636e\u662fN\u548cM\uff0c\u7136\u540e\u662fM\u884c\u6570\u636e\uff0cID K ID[1] ID[2] \u2026 ID[K]\u8868\u793a\u7f16\u53f7\u4e3aID\u7684\u7ed3\u70b9\u6709K\u4e2a\u5b50\u8282\u70b9\uff0c\u5206\u522b\u662fID[1] ID[2] \u2026 ID[K]\u3002\u7ed3\u70b9\u7684\u7f16\u53f7\u662f\u4e24\u4f4d\u6570\u3002<\/p>\n
\u9898\u76ee\u5206\u6790<\/h3>\n
\u8bbeans[x]\u8868\u793a\u7b2cx\u5c42\u7684\u53f6\u5b50\u7ed3\u70b9\u7684\u4e2a\u6570\uff0cnum\u8868\u793a\u6811\u7684\u5c42\u6570\u3002
\n\u904d\u5386\u6bcf\u4e2a\u7ed3\u70b9\uff0c\u5224\u65ad\u662f\u5426\u662f\u53f6\u5b50\u7ed3\u70b9\uff0c\u7136\u540e\u6c42\u51fa\u5b83\u7684\u5c42\u6570\uff0c\u7ef4\u62a4ans\u6570\u7ec4\u5373\u53ef\u3002\u6c42\u5c42\u6570\u53ef\u4ee5\u901a\u8fc7\u7236\u7ed3\u70b9\u4e00\u5c42\u4e00\u5c42\u7684\u5411\u4e0a\u627e\u3002
\n\u7ed3\u70b9\u7684\u5b9a\u4e49\u5982\u4e0b\uff1a<\/p>\n
struct node\n{\n vector<int> children;\n int father;\n} p[105];\n<\/code><\/pre>\n\u4f7f\u7528\u9012\u5f52\u6c42\u5c42\u6570<\/p>\n
int f(int x) \/\/\u8fd4\u56dep[x]\u6240\u5904\u7684\u5c42\u6570\uff0c\u8bbe\u6839\u7ed3\u70b9\u5c42\u6570\u4e3a0\n{\n if (x == 1)\n return 0;\n return f(p[x].father) + 1;\n}\n<\/code><\/pre>\n\u7531\u4e8e\u8f93\u51fa\u65f6\u9700\u8981\u8f93\u51fa\u6bcf\u4e00\u5c42\u7684\u53f6\u5b50\u7ed3\u70b9\u4e2a\u6570\uff0c\u4f46\u6211\u4eec\u5e76\u4e0d\u77e5\u9053\u4e00\u5171\u6709\u591a\u5c11\u5c42\uff0c\u6240\u4ee5\u5728\u904d\u5386\u7ed3\u70b9\u7684\u65f6\u5019\u987a\u4fbf\u7ef4\u62a4\u4e00\u4e0b\u6811\u7684\u5c42\u6570num\u3002
\n\u53e6\u5916\uff0cN=1,M=0\u65f6\u9700\u8981\u8fdb\u884c\u7279\u5224\u3002<\/p>\n
AC\u4ee3\u7801<\/h3>\n#include <bits\/stdc++.h>\nusing namespace std;\nstruct node\n{\n vector<int> children;\n int father;\n} p[105];\nint ans[105] = {0}, num = 0; \/\/\u6bcf\u4e00\u5c42\u7684\u53f6\u5b50\u7ed3\u70b9\u7684\u4e2a\u6570\uff0cnum\u4e3a\u6700\u5927\u5c42\u6570\u7684\u7f16\u53f7\nint f(int x) \/\/\u8fd4\u56dep[x]\u6240\u5904\u7684\u5c42\u6570\uff0c\u8bbe\u6839\u7ed3\u70b9\u5c42\u6570\u4e3a0\n{\n if (x == 1)\n return 0;\n return f(p[x].father) + 1;\n}\nint main()\n{\n int n, m, i, k, id, id2;\n cin >> n >> m;\n if (n == 1 && m == 0) \/\/\u7279\u5224\n {\n cout << 1 << endl;\n return 0;\n }\n while (m--)\n {\n cin >> id >> k;\n while (k--)\n {\n cin >> id2;\n p[id].children.push_back(id2);\n p[id2].father = id;\n }\n }\n for (i = 0; i < 105; i++)\n {\n if (!p[i].children.size() && p[i].father)\n {\n int l = f(i);\n num = max(num, l);\n ans[l]++;\n }\n }\n for (i = 0; i <= num; i++)\n {\n cout << ans[i];\n if (i != num)\n cout << \" \";\n }\n return 0;\n}\n<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"A family hierarchy is usually presented by a pedigree t […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[23,43],"class_list":["post-395","post","type-post","status-publish","format-standard","hentry","category-algo","tag-pat","tag-43"],"_links":{"self":[{"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=\/wp\/v2\/posts\/395","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=395"}],"version-history":[{"count":0,"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=\/wp\/v2\/posts\/395\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=395"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=395"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.guanhaobo.cn\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}