sicily 1034. Forest

1034. Forest

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

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.

Input

There’re several test cases. For each case, in the first line there are two integer numbers n and m (1≤n≤100, 0≤m≤100, m≤n*n) indicating the number of nodes and edges respectively , then m lines followed , for each line of these m lines there are two integer numbers a and b (1≤a,b≤n)indicating there’s an edge pointing from a to b. Nodes are represented by numbers between 1 and n .n=0 indicates end of input.

Output

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.

Sample Input

1 01 11 13 11 32 21 22 10 88

Sample Output

0 1INVALID1 2INVALID

题目分析

求森林最大宽度与深度
拓扑排序,若存在环则失败
每轮确定入度为零的节点数即为该层的宽度
轮数即为深度
一直WA是忽略了题目还限制了不存在两条边指向一个节点,即每个点的入度必须<=1

#include <iostream>#include <vector>#include <memory.h>struct Node {  int indegree;  std::vector<int> child;};int main(){  int num, edge;  while (std::cin >> num >> edge && num != 0) {    Node nodes[num+1];    for (int i = 1; i <= num; ++i) {      nodes[i].indegree = 0;    }    bool visited[num+1];    memset(visited, false, sizeof(visited));    bool valid = true;    int papa, son;    for (int i = 0; i < edge; ++i) {      std::cin >> papa >> son;      nodes[son].indegree++;      nodes[papa].child.push_back(son);      if (nodes[son].indegree == 2)        valid = false;    }    if (!valid) {      std::cout << "INVALID" << std::endl;      continue;    }    int height = -1;    int width = 0;    int count = 0;    while (true) {      std::vector<int> root;      for (int i = 1;  i <= num; ++i) {        if (!visited[i] && nodes[i].indegree == 0) {          root.push_back(i);        }      }      if (root.size() == 0)        break;      width = width < root.size() ? root.size() : width;      for (int i = 0; i < root.size(); ++i) {        visited[root[i]] = true;        count++;        for (int j = 0; j < nodes[root[i]].child.size(); ++j)          nodes[nodes[root[i]].child[j]].indegree--;      }      height++;    }    if (count == num) std::cout << height << " " << width << std::endl;    else std::cout << "INVALID" << std::endl;  }}