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") << "
"; //finish work input_count--; } return 0;}