POJ 1386 Play on Words(判断欧拉回路存在)

//求欧拉回路，注意欧拉图的性质//构图的思想是以单词的首字母和末字母作为结点构图,将单词作为一条有向边//(A)----acm------>(M)-----malform---->(M)------mouse----->(E)//1、第一步是要先对图的连通性进行判断，即去掉边的方向，看图是否连通//我看了数据规模才26，我就偷懒写个Floyd水下算了//2、第二步就是根据欧拉图的性质进行判断就可以了，判断存在欧拉道路或存在欧拉回路//(1)有向图G为欧拉图(存在欧拉回路)，当且仅当G的基图连通，且所有顶点的入度等于出度。//(2)有向图G为半欧拉图(存在欧拉道路)，当且仅当G的基图连通，且存在顶点u的入度比出度大1、v的入度比出度小1，其它所有顶点的入度等于出度。#include<iostream>#include<vector>using namespace std;int indeg[26],outdeg[26],vis[26];int G[26][26];char word[1010];vector<int> letter;int t,m;bool checkEulerPath(){int st,ed;st = ed = 0;bool circle = 1;for(int i = 0;i < letter.size();++i){int x = letter[i];if(indeg[x] - outdeg[x] == 1){++ed;circle = 0;}else if(outdeg[x] - indeg[x] == 1){++st;circle = 0;}else if(outdeg[x] != indeg[x])return false;}if(circle)return true;if(st == 1 && ed == 1)return true;return false;}void Floyd(){for(int k = 0;k < 26;++k)for(int i = 0;i < 26;++i)for(int j = 0;j < 26;++j)if(G[i][k] && G[k][j])G[i][j] = 1;}bool checkConnect(){Floyd();for(int i = 0;i < letter.size();++i)for(int j = 0;j < letter.size();++j){int x = letter[i];int y = letter[j];if(!G[x][y])return false;}return true;}int main(){//freopen("in.txt","r",stdin);scanf("%d",&t);while(t--){scanf("%d",&m);letter.clear();memset(indeg,0,sizeof(indeg));memset(outdeg,0,sizeof(outdeg));memset(G,0,sizeof(G));memset(vis,0,sizeof(vis));while(m--){scanf("%s",word);int last = word[strlen(word) - 1] - 'a';int first = word[0] - 'a';G[first][last] = G[last][first] = 1;indeg[last]++;outdeg[first]++;if(!vis[last]){letter.push_back(last);vis[last] = 1;}if(!vis[first]){letter.push_back(first);vis[first] = 1;}}if(checkConnect() && checkEulerPath())printf("Ordering is possible./n");else printf("The door cannot be opened./n");}return 0;}