二分图解决这个问题的思路
#include <stdio.h>
#include <string>
#include <iostream>
#include <vector>
#include <unordered_map>
#include <queue>
using namespace std;
class Node{
public:
string name;
vector<Node*> adj;
bool visited;
int groupId;
Node(string str):name(str), visited(false), groupId(-1){}
int getNeighborCount() {
return adj.size();
}
};
class Graph{
public:
unordered_map<string, Node*> hashNode;
vector<Node*> nodes;
void createNode(string str)
{
Node* n = new Node(str);
nodes.push_back(n);
hashNode[str] = n;
}
bool isExist(string str)
{
if(hashNode.count(str)>0) return true;
else return false;
}
void addEdge(string str1, string str2)
{
if(!isExist(str1)) createNode(str1);
if(!isExist(str2)) createNode(str2);
hashNode[str1]->adj.push_back( hashNode[str2] );
hashNode[str2]->adj.push_back( hashNode[str1] );
}
bool isBipartite()
{
for(int i=0; i<nodes.size(); i++)
{
if(!nodes[i]->visited)
{
bool f = bfs(nodes[i]);
if(!f) return false;
}
}
return true;
}
bool bfs(Node* n)
{
n->groupId = 1;
queue<Node*> q;
q.push(n);
while(!q.empty())
{
n = q.front(); q.pop();
n->visited = true;
for(int i=0; i<n->getNeighborCount(); i++)
{
if(!n->adj[i]->visited)
{
n->adj[i]->groupId = 3 - n->groupId;
q.push(n->adj[i]);
}
else
{
if(n->adj[i]->groupId == n->groupId)
return false;
}
}
}
return true;
}
~Graph()
{
for(int i=0; i<nodes.size(); i++)
{
delete nodes[i];
}
}
};
int main()
{
freopen("gcj_bad_horse_small-practice-2.in", "r", stdin);
freopen("output.out", "w", stdout);
int input_count;
cin >> input_count;
int counter = 1;
while(input_count)
{
//do work
int M;
cin >> M;
Graph g;
for(int i=0; i<M; i++)
{
string str1, str2;
cin >> str1 >> str2;
g.addEdge(str1, str2);
}
cout << "Case #" << counter++ << ": " << (g.isBipartite()?"Yes":"No") << "\n";
//finish work
input_count--;
}
return 0;
}