POJ2337 Catenyms(欧拉通路的求解)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11526 | Accepted: 2993 |
Description
dog.gopher
gopher.rat
rat.tiger
aloha.aloha
arachnid.dog
A 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.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
Input
Output
Sample Input
2
6
aloha
arachnid
dog
gopher
rat
tiger
3
oak
maple
elm
Sample Output
aloha.arachnid.dog.gopher.rat.tiger
***
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=;
const int M=;
ll power(ll a,int b,ll c) {
ll ans=;
while(b) {
if(b%==) {
ans=(ans*a)%c;
b--;
}
b/=;
a=a*a%c;
}
return ans;
}
struct Word { int l; char s[]; };
struct Edge { int st, ed; bool del; }; Word word[];
Edge edge[];
int in[], out[];
int stk[], father[];
bool mark[];
int E, top; int cmp ( const void* a, const void* b )
{
return strcmp( ((Word*)a)->s, ((Word*)b)->s );
} int find_set ( int x )
{
if ( father[x] != x )
father[x] = find_set ( father[x] );
return father[x];
} bool judge ()
{
int t = ;
for ( int i = ; i < ; i++ )
if ( mark[i] && father[i] == i ) t++;
return t == ;
} void find_path ( int u )
{
for ( int i = ; i < E; i++ )
{
if ( ! edge[i].del && edge[i].st == u )
{
edge[i].del = true;
find_path ( edge[i].ed );
stk[top++] = i;
}
}
} int main()
{
int cs;
scanf("%d",&cs);
while ( cs-- )
{
scanf("%d",&E);
int u, v, c1, c2, start, i; for ( i = ; i < ; i++ )
{
in[i] = out[i] = ;
father[i] = i;
mark[i] = false;
} for ( i = ; i < E; i++ )
{
scanf("%s",word[i].s);
word[i].l = strlen(word[i].s);
} qsort(word, E, sizeof(word[]), cmp); for ( i = ; i < E; i++ )
{
u = word[i].s[] - 'a';
v = word[i].s[word[i].l-] - 'a';
edge[i].st = u;
edge[i].ed = v;
edge[i].del = false;
mark[u] = mark[v] = true;
out[u]++; in[v]++;
u = find_set ( u );
v = find_set ( v );
if ( u != v ) father[v] = u;
} c1 = c2 = ;
start = edge[].st;
for ( i = ; i < ; i++ )
{
if ( in[i] == out[i] ) continue;
else if ( in[i] - == out[i] ) c1++;
else if ( out[i] - == in[i] ) { c2++; start = i; }
else break;
} if ( i == && ((c1 == c2 && c1 == ) || (c1 == c2 && c1 == )) && judge() )
{
top = ;
find_path ( start );
for ( i = top - ; i > ; i-- )
printf("%s.",word[stk[i]].s);
printf("%s\n",word[stk[]].s);
}
else printf("***\n");
}
//system("pause");
return ;
}
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