不相交集数据结构:每棵树的轨迹大小
Disjoint set data structure : track size of each tree
下面是我跟踪不相交集合森林中每棵树的大小的实现。
你能告诉我它有什么问题吗?我正在尝试解决 UVa 问题 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3638
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
class Node {
public :
int id;
Node *parent;
unsigned long long rank;
Node(int id) {
this->id = id;
// this->data = data;
this->rank =1; //size here
this->parent = this;
}
friend class DisjointSet;
};
class DisjointSet {
unordered_map<int,Node*> nodesMap;
Node *find_set_helper(Node *aNode) {
if (aNode == aNode->parent) {
return aNode->parent;
}
return find_set_helper(aNode->parent);
}
void link(Node *xNode,Node *yNode) {
if( xNode->rank > yNode->rank) {
yNode->parent = xNode;
xNode->rank += yNode->rank;
}
// else if(xNode-> rank < yNode->rank){
// xNode->parent = yNode;
// yNode->rank += xNode->rank;
// }
else {
xNode->parent = yNode;
yNode->rank += xNode->rank;
}
}
public:
DisjointSet() {
}
void AddElements(int sz) {
for(int i=0;i<sz;i++)
this->make_set(i);
}
void make_set(int id) {
Node *aNode = new Node(id);
this->nodesMap.insert(make_pair(id,aNode));
}
void Union(int xId, int yId) {
Node *xNode = find_set(xId);
Node *yNode = find_set(yId);
if(xNode && yNode)
link(xNode,yNode);
}
Node* find_set(int id) {
unordered_map<int,Node*> :: iterator itr = this->nodesMap.find(id);
if(itr == this->nodesMap.end())
return NULL;
return this->find_set_helper(itr->second);
}
~DisjointSet(){
unordered_map<int,Node*>::iterator itr;
for(itr = nodesMap.begin(); itr != nodesMap.end(); itr++) {
delete (itr->second);
}
}
};
int main() {
int n,m,k,first,cur;
//freopen("in.in","r",stdin);
scanf("%d %d",&n,&m);
while(n != 0 || m != 0) {
DisjointSet *ds = new DisjointSet();
ds->AddElements(n); // 0 to n-1
//printf("\n n = %d m = %d",n,m);
for(int i=1;i<=m;i++) {
scanf("%d",&k);
//printf("\nk=%d",k);
if ( k > 0 ) {
scanf("%d",&first);
for(int j=2;j<=k;j++) {
scanf("%d",&cur);
ds->Union(first,cur);
}
}
}
Node *zeroSet = ds->find_set(0);
// unsigned long long count = ds->getCount(zeroSet->id);
printf("%llu\n",zeroSet->rank);
delete ds;
scanf("%d %d",&n,&m);
}
return 0;
}
上面代码中的 link 函数完成更新树大小的工作。
问题的解决方法是找到元素0所属的集合,并得到集合中代表元素的大小。
但是我用这段代码得到了错误的答案。
你能帮帮我吗
在你的 Union
函数中,检查两个节点是否已经在同一个集合中。
if(xNode && yNode && xNode != yNode)
link(xNode,yNode);
下面是我跟踪不相交集合森林中每棵树的大小的实现。
你能告诉我它有什么问题吗?我正在尝试解决 UVa 问题 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3638
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
class Node {
public :
int id;
Node *parent;
unsigned long long rank;
Node(int id) {
this->id = id;
// this->data = data;
this->rank =1; //size here
this->parent = this;
}
friend class DisjointSet;
};
class DisjointSet {
unordered_map<int,Node*> nodesMap;
Node *find_set_helper(Node *aNode) {
if (aNode == aNode->parent) {
return aNode->parent;
}
return find_set_helper(aNode->parent);
}
void link(Node *xNode,Node *yNode) {
if( xNode->rank > yNode->rank) {
yNode->parent = xNode;
xNode->rank += yNode->rank;
}
// else if(xNode-> rank < yNode->rank){
// xNode->parent = yNode;
// yNode->rank += xNode->rank;
// }
else {
xNode->parent = yNode;
yNode->rank += xNode->rank;
}
}
public:
DisjointSet() {
}
void AddElements(int sz) {
for(int i=0;i<sz;i++)
this->make_set(i);
}
void make_set(int id) {
Node *aNode = new Node(id);
this->nodesMap.insert(make_pair(id,aNode));
}
void Union(int xId, int yId) {
Node *xNode = find_set(xId);
Node *yNode = find_set(yId);
if(xNode && yNode)
link(xNode,yNode);
}
Node* find_set(int id) {
unordered_map<int,Node*> :: iterator itr = this->nodesMap.find(id);
if(itr == this->nodesMap.end())
return NULL;
return this->find_set_helper(itr->second);
}
~DisjointSet(){
unordered_map<int,Node*>::iterator itr;
for(itr = nodesMap.begin(); itr != nodesMap.end(); itr++) {
delete (itr->second);
}
}
};
int main() {
int n,m,k,first,cur;
//freopen("in.in","r",stdin);
scanf("%d %d",&n,&m);
while(n != 0 || m != 0) {
DisjointSet *ds = new DisjointSet();
ds->AddElements(n); // 0 to n-1
//printf("\n n = %d m = %d",n,m);
for(int i=1;i<=m;i++) {
scanf("%d",&k);
//printf("\nk=%d",k);
if ( k > 0 ) {
scanf("%d",&first);
for(int j=2;j<=k;j++) {
scanf("%d",&cur);
ds->Union(first,cur);
}
}
}
Node *zeroSet = ds->find_set(0);
// unsigned long long count = ds->getCount(zeroSet->id);
printf("%llu\n",zeroSet->rank);
delete ds;
scanf("%d %d",&n,&m);
}
return 0;
}
上面代码中的 link 函数完成更新树大小的工作。
问题的解决方法是找到元素0所属的集合,并得到集合中代表元素的大小。 但是我用这段代码得到了错误的答案。
你能帮帮我吗
在你的 Union
函数中,检查两个节点是否已经在同一个集合中。
if(xNode && yNode && xNode != yNode)
link(xNode,yNode);