poj 1386 Play on Words

转换成图论一笔画问题

如何转换，一个单词的第一个字母和最后一个字母作为点，由第一个字母向最后一个字母连一条有向边

问题转化为判断一个有向图是否存在欧拉回路或者欧拉路径

有向图判断是否存在欧拉回路或者欧拉路径首先要满足其基图（即把边改成无向的图）是连通的（并查集判断）

判断的时候要注意两种情况回路或者路径，满足回路的条件是每个定点的入度与出度相等，路径的条件是当且仅有一个顶点

的入度比出度大一，和当且仅有一个顶点的出度比入度大一

#include<stdio.h>#include<string.h>const int MAXN = 26;int du[MAXN][2];//du[0] 入度 d[1] 出度int father[MAXN];bool used[MAXN];void makeSet(){for(int i = 0; i < MAXN; ++i)father[i] = i;}int find(int x){return father[x] == x ? father[x] : find(father[x]);}void merge(int x, int y){father[x] = father[y];}bool judge(){int more = 0, less = 0, i;for(i = 0; i < MAXN; ++i){if(du[i][0] == du[i][1] || !used[i])continue;if(du[i][0] - du[i][1] == 1){more++;} else if(du[i][1] - du[i][0] == 1){less++;} else {break;}}if(i < MAXN || more != less || (more == less && more > 1))//判断两种情况return false;int s;//并查集判断基图是否连通for(i = 0; i < MAXN; ++i){if(used[i]){s = i;break;}}int root = find(s);for(i = 0; i < MAXN; ++i){if(used[i] && find(i) != root)return false;}return true;}int main(){int t, n;char word[1001];scanf("%d",&t);while(t--){memset(du, 0, sizeof(du));memset(used, 0, sizeof(used));makeSet();scanf("%d",&n);for(int i = 0; i < n; ++i){scanf("%s",word);int len = strlen(word);used[word[0] - 'a'] = used[word[len - 1] - 'a'] = true;du[word[0] - 'a'][1]++;du[word[len - 1] - 'a'][0]++;int x = find(word[0] - 'a'), y = find(word[len - 1] - 'a');if(x != y)//合并merge(x, y);}if(judge())printf("Ordering is possible./n");elseprintf("The door cannot be opened./n");}return 0;}