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sicily 1034. Forest
In the field of computer science, forest is important and deeply researched , it is a model for many data structures . Now it’s your job here to calculate the depth and width of given forests.
Precisely, a forest here is a directed graph with neither loop nor two edges pointing to the same node. Nodes with no edge pointing to are roots, we define that roots are at level 0 . If there’s an edge points from node A to node B , then node B is called a child of node A , and we define that B is at level (k+1) if and only if A is at level k .
We define the depth of a forest is the maximum level number of all the nodes , the width of a forest is the maximum number of nodes at the same level.For each case output one line of answer , if it’s not a forest , i.e. there’s at least one loop or two edges pointing to the same node, output “INVALID”(without quotation mark), otherwise output the depth and width of the forest, separated by a white space.
1 01 11 13 11 32 21 22 10 88
0 1INVALID1 2INVALID
这题主要考虑环的问题吧,还有要注意的就是单个元素自身成环和没有根节点的情况
#include <iostream>#include <cstring>#include <stack>#include <vector>using namespace std;struct Node{ Node(int v = -1, int l = 0):value(v), level(l) { } int value; int level;};int main(int argc, char const *argv[]){ int n, m, a, b; int level[101]; // 记录每层的个数 bool visited[101]; // 用于判断是否形成环 while (cin >> n >> m && n != 0) { vector<int> node[101]; Node parent[101]; // 记录父节点 memset(level, 0, sizeof(level)); memset(visited, false, sizeof(visited)); bool isSucceed = true; for (int i = 0; i != m; ++i) { cin >> a >> b; if (a == b) { // 元素指向自身 isSucceed = false; continue; } node[a - 1].push_back(b - 1); // 元素值是从 1 开始的 if (parent[b - 1].value =http://www.mamicode.com/= -1) parent[b - 1].value = http://www.mamicode.com/a - 1; else isSucceed = false; } bool hasRoot = false; // 是否有根节点 if (isSucceed) { for (int i = 0; i != n; ++i) { if (!isSucceed) break; if (parent[i].value =http://www.mamicode.com/= -1) { // 从根节点遍历每一棵树 hasRoot = true; stack<int> s; s.push(i); while (!s.empty()) { if (!isSucceed) break; int current = s.top(); visited[current] = true; if (current != i) parent[current].level = parent[parent[current].value].level + 1; // 子节点level等于父节点+1 s.pop(); for (int j = 0; j != node[current].size(); ++j) { if (visited[node[current][j]]) { // 形成了环 isSucceed = false; break; } s.push(node[current][j]); } } } } } if (isSucceed) { for (int i = 0; i != n; ++i) { level[parent[i].level]++; } } if (!isSucceed || !hasRoot) { cout << "INVALID" << endl; continue; } int levelMax = -1; int index = -1; for (int i = 0; i != n; ++i) { if (level[i] == 0) break; if (levelMax < level[i]) { levelMax = level[i]; } index++; } cout << index << " " << levelMax << endl; } return 0;}
sicily 1034. Forest