pku 2337 Catenyms(寻找欧拉通路)

以字母为节点，字符串为边，寻找欧拉回路。

一开始不会，用的深搜回溯找的路径。后来问了lvyun，才知道有个很好用的套圈法,每条边只要走一次O(e)的效率。

关于套圈法，可以参看http://blog.csdn.net/logic_nut/archive/2009/08/23/4474307.aspx

深搜回溯

#include <iostream>using namespace std;int degree_in[26],degree_out[26],edge_num,edge_order[1005];bool used[1005];struct Edge{int start,end;char word[25];}edges[1005];int CMP(const void* a,const void* b){Edge *c=(Edge*)a,*d=(Edge*)b;if(c->start!=d->start) return c->start-d->start;else return strcmp(c->word,d->word);}inline int ABS(int a){if(a>0) return a;else return -a;}int check()//返回起点{int cnt1=0,cnt2=0,source;for(int i=0;i<26;i++){if(ABS(degree_in[i]-degree_out[i])==2) return -1;else if(degree_in[i]-degree_out[i]==1){cnt1++;//基度顶点计数 }else if(degree_in[i]-degree_out[i]==-1){cnt2++;//基度顶点计数 source=i;}}if(cnt1>1||cnt2>1) return -1;else if(cnt1==0){for(int i=0;i<26;i++)if(degree_out[i]) return i;}else return source;}bool solve_dfs(int source,int cnt){if(cnt==edge_num) return true;for(int i=0;i<edge_num;i++){if(edges[i].start<source||used[i]) continue;else if(edges[i].start>source) return false;used[i]=true;edge_order[cnt]=i;if(solve_dfs(edges[i].end,cnt+1)) return true;used[i]=false;}return false;}int main(){int t;scanf("%d",&t);while(t--){memset(degree_out,0,sizeof(degree_out));memset(degree_in,0,sizeof(degree_in));scanf("%d",&edge_num);for(int i=0;i<edge_num;i++){scanf("%s",edges[i].word);edges[i].start=edges[i].word[0]-'a';edges[i].end=edges[i].word[strlen(edges[i].word)-1]-'a';degree_out[edges[i].start]++;degree_in[edges[i].end]++;}int source=check();if(source==-1){printf("***/n");continue;}qsort(edges,edge_num,sizeof(edges[0]),CMP);memset(used,0,sizeof(used));if(!solve_dfs(source,0)){printf("***/n");continue;}printf("%s",edges[edge_order[0]].word);for(int i=1;i<edge_num;i++){printf(".%s",edges[edge_order[i]].word);}printf("/n");}}

 

套圈法（three_up队lvyun的代码）

#include <iostream>using namespace std;struct adjLink{int v, nxt; char word[30]; }adj[2000]; int adjN; void insert(int beg, int end, char word[30])//类似于hash，第一个节点数据域不存值只有指针部分有效。链表中单词按字典序排列。{ int i; for (i = beg; adj[i].nxt != -1; i = adj[i].nxt)if (strcmp(adj[adj[i].nxt].word, word) >= 0) break; adj[adjN].v = end; strcpy(adj[adjN].word, word); adj[adjN].nxt = adj[i].nxt; adj[i].nxt = adjN++; }int vis[30], count; void dfs(int cur)//检查连通性{ vis[cur] = 1; count++; for (int i = adj[cur].nxt; i != -1; i = adj[i].nxt)if (!vis[adj[i].v]) dfs(adj[i].v); }int path[2000], npath; void euler(int cur, int edgeN) //cur当前到达的节点 edgeN上一被选择的边，即上一个节点通过edgeN到达的cur{ int i;while(adj[cur].nxt != -1){i=adj[cur].nxt;adj[cur].nxt=adj[i].nxt;euler(adj[i].v,i);} path[npath++] = edgeN; //后序记录 输出的时候为了保证字典序，需逆向输出} void go(){ int n, i; scanf("%d", &n); for (i = 0; i < 26; i++) adj[i].nxt = -1; adjN = i; int flag[30], tot = 0, cnt[30]; memset(flag, 0, sizeof(flag)); memset(cnt, 0, sizeof(cnt)); for (i = 0; i < n; i++) { char word[30];scanf("%s", &word);int a = word[0] - 'a', b = word[strlen(word) - 1] - 'a'; int len = strlen(word); word[len] = '.'; word[len + 1] = 0; insert(a, b, word); cnt[b]--; cnt[a]++; if (!flag[a])flag[a] = 1, tot++; if (!flag[b]) flag[b] = 1, tot++; }for (i = 0; i < 26; i++) { memset(vis, 0, sizeof(vis)); count = 0; dfs(i); if (count == tot) break; }if (i == 26){ printf("***/n"); return ; }int flagIn = 0, flagOut = 0, flagNo = 0, site; for (i = 0; i < 26; i++) if (cnt[i] == -1) flagIn++; else if (cnt[i] == 1) flagOut++, site = i; elseif (cnt[i]) flagNo = 1; if (flagNo || flagIn > 1 || flagOut > 1 || flagIn != flagOut)//根据欧拉通路的充要条件 将不符合的先排除掉{ printf("***/n"); return ; }npath = 0; if (flagOut == 0){ for (i = 0; i < 26; i++) if (adj[i].nxt != -1){site = i; break; }} euler(site, -1); //因为直接从site出发，所以第二个参数用-1代替，输出的时候要忽略掉。for (i = npath - 2; i > 0; i--) printf("%s", adj[path[i]].word); int len = strlen(adj[path[0]].word);adj[path[0]].word[len - 1] = 0; printf("%s/n", adj[path[0]].word); }int main(){ int T; scanf("%d", &T); while (T--)go(); return 0; }