PAT 1021

1021. Deepest Root (25)

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes' numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print "Error: K components" where K is the number of connected components in the graph.

Sample Input 1:

Sample Output 1:

3
4
5

Sample Input 2:

Sample Output 2:

Error: 2 components

首先说说算法的思路：先用并查集判断树是否是联通；然后任选一个结点做bfs，则得到最低层的叶结点一定是Deepest Root，接下去在从已选出的Deepest Root中任选一点，在做一次dfs，然后在
将最底层的叶结点加入Deepest Root。此题在时间和空间上都有限制，如果暴力解决的话会超时，如果用邻接矩阵存树的话会超内存，所以采用链表存树。
代码

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <stdlib.h>
  4 
  5 #define NUM 10001
  6 typedef struct linkNode{
  7     int a;
  8     linkNode *next;
  9 }linkNode;
 10 
 11 linkNode *map[NUM];
 12 int unionSet[NUM];
 13 int level[NUM];
 14 int queue[NUM];
 15 int deepestRoot[NUM];
 16 
 17 int findRoot(int);
 18 int bfs(int);
 19 void destroyMap(int);
 20 
 21 int main()
 22 {
 23     int N,i;
 24     int s,e;
 25     int k;
 26     linkNode *p;
 27     while(scanf("%d",&N) != EOF){
 28         memset(unionSet,0,sizeof(unionSet));
 29         memset(map,0,sizeof(map));
 30         k = 0;
 31         for(i=1;i<N;++i){
 32             scanf("%d%d",&s,&e);
 33             int sR = findRoot(s);
 34             int eR = findRoot(e);
 35             if(sR != eR){
 36                 unionSet[sR] = eR;
 37                 ++k;
 38                 p = (linkNode*) malloc(sizeof(linkNode));
 39                 p->a = e;
 40                 p->next = NULL;
 41                 if(map[s]){
 42                     p->next = map[s];
 43                     map[s] = p;
 44                 }
 45                 else
 46                     map[s] = p;
 47                 p = (linkNode*) malloc(sizeof(linkNode));
 48                 p->a = s;
 49                 p->next = NULL;
 50                 if(map[e]){
 51                     p->next = map[e];
 52                     map[e] = p;
 53                 }
 54                 else
 55                     map[e] = p;
 56             }
 57         }
 58         if(k < N - 1){
 59             printf("Error: %d components
",N-k);
 60             destroyMap(N);
 61             continue;
 62         }
 63         memset(deepestRoot,0,sizeof(deepestRoot));
 64         int maxLevel = bfs(1);
 65         int maxLevel_i;
 66         for(i=1;i<=N;++i){
 67             if(level[i] == maxLevel){
 68                 deepestRoot[i] = 1;
 69                 maxLevel_i = i;
 70             }
 71         }
 72         maxLevel = bfs(maxLevel_i);
 73         for(i=1;i<=N;++i){
 74             if(level[i] == maxLevel)
 75                 deepestRoot[i] = 1;
 76         }
 77         for(i=1;i<=N;++i){
 78             if(deepestRoot[i])
 79                 printf("%d
",i);
 80         }
 81         destroyMap(N);
 82     }
 83     return 0;
 84 }
 85 
 86 int findRoot(int s)
 87 {
 88     while(unionSet[s])
 89         s = unionSet[s];
 90     return s;
 91 }
 92 
 93 
 94 int bfs(int s)
 95 {
 96     memset(level,-1,sizeof(level));
 97     int base = 0,top = 0;
 98     int levelNum = 1,endLevel = 1;
 99     int t,i;
100     linkNode *p;
101     queue[top++] = s;
102     while(top > base){
103         t = queue[base++];
104         level[t] = levelNum;
105         p = map[t];
106         while(p){
107             if(level[p->a] == -1){
108                 queue[top++] = p->a;
109             }
110             p = p->next;
111         }
112         if(endLevel == base){
113             endLevel = top;
114             ++levelNum;
115         }
116     }
117     return levelNum - 1;
118 }
119 
120 void destroyMap(int n)
121 {
122     linkNode *p,*q;
123     int i;
124     for(i=1;i<=n;++i){
125         p = map[i];
126         while(p){
127             q = p->next;
128             free(p);
129             p = q;
130         }
131     }
132 }

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原文地址：https://www.cnblogs.com/boostable/p/pat_2021.html