Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
| Total Submissions: 30515 | Accepted: 8056 |
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.
先不說這題有多坑 先說這題的題意：
給一些棒子，每根棒子的兩頭有顏色，用兩個字符串給出。問能不能將這些棒子首尾相接，使得首尾的顏色都相同。
剛開始想的非常簡單，直接用map做個映射然後判斷有沒有歐拉路徑（兩個或者零個度爲奇數的點）
直接TLE了，沒辦法，改用trie樹存字符串，然後光榮的WA了
後來走投無路看了網上題解，說是有不連通的情況。好吧 是我自己的問題，果斷加了一段並查集。
最坑的地方出現了 ，居然還是wa了
然後就是一段漫長的交題過程。。。。
中午回去睡了一覺，下午來看，想了想，加了一組什麼都沒有的特判（其實就是亂加了）
結果過了。
過了。。
說不下去了。。太難過了。。。
代碼：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int trie[750005][30];
int ok[750005];
int fa[750005];
int id[750005];
int sz,num;
int getfa(int x){
return fa[x] != x? fa[x] = getfa(fa[x]) : x;
}
void init(){
for(int i=0;i<250005;i++)fa[i] = i;
memset(trie,0,sizeof(trie));
memset(ok,0,sizeof(ok));
memset(id,0,sizeof(id));
sz = 1;
num = 0;
}
int in(char *s) {
int cur = 1;
for (int i= 0; s[i]; i++) {
int c = s[i]-'a';
if (!trie[cur][c]){
trie[cur][c]= ++sz;
}
cur = trie[cur][c];
}ok[cur]++;
if(ok[cur]==1){id[cur] = ++num;return num;}
return id[cur];
}
int main()
{
char str1[15],str2[15];
init();
while(scanf("%s%s",str1,str2)==2){
int a,b;
a = in(str1);
b = in(str2);
fa[getfa(a)] = getfa(b);
}
if(num == 0){cout<<"Possible"<<endl;return 0;}
int flag,cnt =0,cnt2 = 0;
for(int i=1;i<=sz;i++){
if(ok[i]%2!=0)cnt++;
if(cnt>2)break;
}
for(int i=1;i<=num;i++)if(fa[i] == i)cnt2++;
if(cnt2 == 1&&(cnt == 2|| cnt ==0))flag = 1;
else flag = 0;
printf("%s\n",flag?"Possible":"Impossible");
return 0;
}