joj 1857 Catenyms
1857: Catenyms
| Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
|---|---|---|---|---|---|
| 3s | 8192K | 84 | 17 | Standard |
dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dogA compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tigerGiven a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once. The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself. For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.
Sample Input
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm
Output for Sample Input
aloha.arachnid.dog.gopher.rat.tiger ***
This problem is used for contest: 171
//无语了 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Edge { int u,v,next; char str[31]; } edge[1010]; bool cmp(Edge a,Edge b) { if(strcmp(a.str,b.str)>=0) return 1; else return 0; } int ttt; int od[31],id[31]; bool buse[31]; int p[31]; int head[31]; bool vis[2010]; int list[1010]; int n; int num; void ufset() { for(int i=0; i<26; i++) p[i]=i; } int find(int x) { if(x!=p[x]) p[x]=find(p[x]); return p[x]; } void Union(int r1,int r2) { int t1=find(r1); int t2=find(r2); p[t1]=t2; } bool bsconnect() { int u,v; ufset(); int i; for(i=0; i<n; i++) { u=edge[i].u; v=edge[i].v; if(u!=v && find(u)!=find(v)) Union(u,v); } int first=-1; for(i=0; i<26; i++) { if(buse[i]==0) continue; if(first==-1) first=i; else if(find(i)!=find(first)) break; } if(i<26) return false; else return true; } void dfs(int x,int cas) { for(int p=head[x]; p!=-1; p=edge[p].next) if(!vis[p]) { vis[p]=1; dfs(edge[p].v,p); } if(cas!=-1) list[num++]=cas; } void output() { int u,v; for(int i=num-1;i>0;i--) printf("%s.",edge[list[i]].str); printf("%s/n",edge[list[0]].str); } int main() { int T; int u,v; scanf("%d",&T); while(T--) { memset(od,0,sizeof(od)); memset(id,0,sizeof(id)); memset(buse,0,sizeof(buse)); memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%s",edge[i].str); u=edge[i].str[0]-'a'; v=edge[i].str[strlen(edge[i].str)-1]-'a'; od[u]++; id[v]++; buse[u]=buse[v]=1; edge[i].u=u; edge[i].v=v; } bool euler=true; int one=0; int none=0; int tt; for(int j=0; j<26; j++) { if(buse[j]==0) continue; if(od[j]-id[j]>=2||id[j]-od[j]>=2) { euler=false; break; } if(od[j]==0&&id[j]==0) { euler=false; break; } if(od[j]-id[j]==1) { one++; tt=j; if(one>1) { euler=false; break; } } if(id[j]-od[j]==1) { none++; if(none>1) { euler=false; break; } } } if(one!=none) euler=false; if(!bsconnect()) euler=false; if(euler) { sort(edge,edge+n,cmp); for(int i=0; i<n; i++) { u=edge[i].u; edge[i].next=head[u]; head[u]=i; } num=0; if(one==0&&none==0) { for(int i=0; i<26; i++) if(buse[i]) { dfs(i,-1); break; } } else { dfs(tt,-1); } output(); } else printf("***/n"); } return 0; }