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
思路
題意很多棍子，顏色相同的兩端可以拼接在一起，問所有的棍子能不能拼成一根長棍子。emmm…歐拉路問題，只不過要對這個單詞做一個hash吧。然後hash的方式比較多，反正比較容易想到的就是map，unorder_map(哈希表)，字典樹。北大OJ真的是太老了，map超時可能是我手法問題？unorder_map不支持會報編譯錯誤…只能硬着頭皮寫字典樹手動hash了。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
const int maxn = 500005;
int trie[maxn][26];
int num[maxn];
int s[maxn];
int in[maxn];
int cnt,res;
int height[maxn]; //按高度合併降一點複雜度，也可以不寫。
void clear_set()
{
res = cnt = 0;
memset(height,0,sizeof(height));
memset(in,0,sizeof(in));
for(int i = 0;i < maxn;i++){
s[i] = i;
}
}
int insert_s(char *s)
{
int n = strlen(s);
int root = 0;
for(int i = 0;i < n;i++){
int k = s[i] - 'a';
if(!trie[root][k]){
trie[root][k] = ++cnt;
}
root = trie[root][k];
}
if(!num[root]){
num[root] = ++res;
}
return num[root];
}
int find_set(int x)
{
if(s[x] != x){
s[x] = find_set(s[x]);
}
return s[x];
}
void union_set(int x,int y)
{
int fx = find_set(x);
int fy = find_set(y);
if(fx == fy) return ;
if(height[fx] == height[fy]){
s[fy] = fx;
height[fx]++;
}
else{
if(height[fx] < height[fy]){
s[fx] = fy;
}
else{
s[fy] = fx;
}
}
}
int main()
{
char t[30],p[30];
clear_set();
while(scanf("%s%s",t,p)!= EOF){
int x = insert_s(t);
int y = insert_s(p);
in[x]++;in[y]++;
union_set(x,y);
}
int sum = 0,ans = 0;
for(int i = 1;i <= res;i++){
if(s[i] == i){
ans++;
}
if(in[i]%2){
sum++;
}
}
if(ans > 1){
printf("Impossible\n");
}
else{
if(sum ==2 || sum == 0){
printf("Possible\n");
}
else{
printf("Impossible\n");
}
}
return 0;
}
願你走出半生，歸來仍是少年~