标签：des style blog color io os ar for sp
Time Limit: 1 secs, Memory Limit: 32 MB
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.
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.
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 0 1 1 1 1 3 1 1 3 2 2 1 2 2 1 0 88
0 1 INVALID 1 2 INVALID
分析：一看到题目中说到了有环的情况，立马就想到了学数据结构的时候用广搜来进行拓扑排序可以找出环。这道题同样也是用广搜，通过边遍历每一个可以放入森林的节点，每次遍历到一个节点，就删掉指向它的边，要是存在某些边没有被删除，就说明有些节点是非法的。每个节点都有一个level来表示它所在的层数，子节点的层数等于父节点层数加1。
1 #include <iostream> 2 #include <memory.h> 3 #include <queue> 4 #include <vector> 5 using namespace std; 6 7 struct Node{ 8 int id; //树中每个节点的编号 9 int level; //树中每个节点所在的层数 10 }; 11 12 struct Edge{ 13 int from; 14 int to; 15 }; 16 17 int main(){ 18 int n, m, i; 19 Edge edge; 20 Node node; 21 Node nodes[100]; 22 vector<Edge>::iterator iter; 23 bool isRoot[100]; 24 while(cin >> n >> m && n != 0){ 25 vector<Edge> v; 26 queue<Node> q; 27 memset(isRoot, true, sizeof(isRoot)); 28 for(i = 0; i < n; i++){ 29 nodes[i].id = i + 1; 30 nodes[i].level = -1; 31 } 32 for(i = 0; i < m; i++){ 33 cin >> edge.from >> edge.to; 34 v.push_back(edge); 35 isRoot[edge.to - 1] = false; 36 } 37 for(i = 0; i < n; i++){ 38 if(isRoot[i]){ 39 node.id = i + 1; 40 node.level = 0; 41 q.push(node); 42 } 43 } 44 /*广搜来标记每个节点的层数*/ 45 while(!q.empty()){ 46 node = q.front(); 47 q.pop(); 48 for(iter = v.begin(); iter != v.end(); iter++){ 49 if(iter->from == node.id && nodes[iter->to -1].level == -1){ 50 nodes[iter->to -1].level = node.level+1; 51 q.push(nodes[iter->to -1]); 52 v.erase(iter); //每次找到一个合法的点就删掉一条边 53 iter--; 54 } 55 } 56 } 57 if(!v.empty()){ //有环或者存在几个父节点同时指向一个子节点 58 cout << "INVALID" << endl; 59 } else { 60 int sum[100]; //标记每层节点的个数 61 int max = 0; 62 memset(sum, 0, sizeof(sum)); 63 for(i = 0; i < n; i++){ 64 if(nodes[i].level == -1){ 65 nodes[i].level = 0; 66 } 67 sum[nodes[i].level]++; 68 } 69 for(i = 0; i < 100; i++){ 70 if(sum[i]==0){ 71 break; 72 } 73 if(max < sum[i]){ 74 max = sum[i]; 75 } 76 } 77 cout << i - 1 << " " << max << endl; 78 } 79 } 80 return 0; 81 }
附加测试样例：
输入：4 2
1 2
3 4
输出：1 2
标签：des style blog color io os ar for sp
原文地址：http://www.cnblogs.com/hunter-lee/p/4032378.html
