Luogu p Rui's wooden stick, P1333 wooden stick

Source: Internet
Author: User

Luogu p Rui's wooden stick, P1333 wooden stick
Description

Rui has a bunch of toy wooden sticks, each of which has been infected with a certain color. Now he suddenly has an idea that he wants to splice these wooden sticks together into a line, in addition, the color of the two ends of the contact with the wooden stick is the same. The color of the two ends of each wooden stick is given. Is there any arrangement that meets the requirements.

For example, if there are only two wooden sticks, the colors at both ends are red and blue, and the colors at both ends are red and yellow, blue --- red | red ---- yellow is a suitable arrangement.

Input/Output Format

Input Format:

There are several lines in the input. Each line contains two words, indicating the color at both ends of a wooden stick. The word is composed of lowercase letters, and the word length cannot exceed 10 letters. A maximum of 250000 wooden sticks can be entered.

Output Format:

If the sticks can be arranged as required, output Possible; otherwise, output Impossible

Input and Output sample input sample #1: Copy
`blue redred violetcyan blueblue magentamagenta cyan`
Output example #1: Copy
`Possible`

Let's look at the dots of the same color as a node.

Then this question is to determine whether there is an Euler loop.

The Euler's loop is basically determined by DFS.

But in fact, you can also query the set.

Set \$ x \$ to the number of successful merge operations, and \$ n \$ to the number of points.

Then the whole graph is an Euler loop. if and only when \$ x >n-1 and the odd degree period is \$0 \$ or \$2 \$

By the way

Pbds is really a good thing.

`#include<cstdio>#include<cstring>#include<map>#include<iostream>#include<ext/pb_ds/assoc_container.hpp>#include<ext/pb_ds/hash_policy.hpp>using namespace __gnu_pbds;using namespace std;const int MAXN=1e6+10;gp_hash_table<string,int>mp;int tot=0,fa[MAXN],inder[MAXN];int find(int x){    if(fa[x]==x) return fa[x];    else return fa[x]=find(fa[x]);}int unionn(int x,int y){    inder[x]++;inder[y]++;    int fx=find(x),fy=find(y);    if(fx==fy) return 0;    fa[fx]=fy; return 1;}int main(){    ios::sync_with_stdio(false);     for(int i=1;i<=250001;i++) fa[i]=i;    string a,b;    int ans=0;    while(cin>>a>>b)    {        int posa=mp[a]?mp[a]:mp[a]=++tot;        int posb=mp[b]?mp[b]:mp[b]=++tot;        ans+=unionn(posa,posb);        }    if(ans<tot-1) {printf("Impossible\n");return 0;}    int attack=0;    for(int i=1;i<=tot;i++)        if(inder[i]&1) attack++;     if(attack>2) {printf("Impossible\n");return 0;}    printf("Possible\n");    return 0;}`

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