Time Limit: 5000MS Memory Limit: 128000K
Total Submissions: 13696 Accepted: 3384
Description
You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a straight line such that the colors of the endpoints that touch are of the same color?
Input
Input is a sequence of lines, each line contains two words, separated by spaces, giving the colors of the endpoints of one stick. A word is a sequence of lowercase letters no longer than 10 characters. There is no more than 250000 sticks.
Output
If the sticks can be aligned in the desired way, output a single line saying Possible, otherwise output Impossible.
Sample Input
blue red
red violet
cyan blue
blue magenta
magenta cyan
Sample Output
Possible
Hint
Huge input,scanf is recommended.
Source
The UofA Local 2000.10.14
//
#include <iostream>
using namespace std;
//Tries
struct Trie
{
Trie():id(-1),end(false){memset(next, 0, sizeof(next));}
int id;
bool end;
Trie* next[26];
};
static int num = 0;
int GetID(char *x, Trie* root)
{
Trie *temp = root;
for (int i = 0; i < strlen(x); ++i)
{
if (temp->next[x[i]-'a'] == NULL)
temp->next[x[i]-'a'] = new Trie;
temp = temp->next[x[i]-'a'];
}
if (temp->end)return temp->id;
temp->end = true;
temp->id = num++;
return temp->id;
}
//Disjoint set
int Find(int x, int f[])
{
if(x != f[x])
return f[x] = Find(f[x], f);
return f[x];
};
void Union(int x,int y, int f[])
{
f[Find(x, f)] = f[Find(y, f)];
};
int main(int argc, char* argv[])
{
Trie root;
char w[30], w1[12], w2[12];
int degree[500001];
memset(degree,0,sizeof(degree));
int f[500001];
for(int i = 0; i < 500001; ++i) f[i] = i;
//create Trie tree
while (gets(w) && w[0] != 0)
{
sscanf(w,"%s %s", w1, w2);
int x = GetID(w1, &root);
int y = GetID(w2, &root);
++degree[x];
++degree[y];
Union(x,y,f);
}
//check odd points
int odd = 0;
for (int i = 0; i < num; ++i)
if (degree[i] & 1 != 0) ++odd;
//check connected
int x = Find(1,f);
for (int i = 0; i < num ; ++i)
if (x != Find(i,f))
{
odd = -1;
break;
};
if (odd == 0 || odd == 2) cout << "Possible" << endl;
else cout << "Impossible" << endl;
return 0;
}