poj 2513 Colored Sticks（字典树+欧拉路径+并查集）

最新推荐文章于 2019-10-30 17:17:00 发布

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解决一个有趣的问题：如何判断一组两端涂有不同颜色的木棍是否可以首尾相连形成一条直线，确保相邻木棍接触的两端颜色一致。通过构建无向图，并利用并查集与字典树来高效处理大量数据。

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 27267		Accepted: 7212

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.

题意：给出n条线段，线段的两个端点都有颜色。线段两两相连当且仅当相连的端点颜色相同，问能否将所有线段连成一条。

思路：判断是否存在欧拉路径。

无向图存在欧拉路的充要条件为：

① 图是连通的；（用并查集判断）

② 所有节点的度为偶数，或者有且只有两个度为奇数的节点

另外，需要将所有字符串插入到字典树中，这样能快速找到字符串对应的下标。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;

const int maxn=250005*2;

typedef struct TireNode
{
    int id;
    struct TireNode *next[26];
}TireNode;
TireNode Memory[1000000];
int allocp=0;
TireNode *root;
int fa[maxn],degree[maxn];
TireNode *CreateTireNode()
{
    TireNode *p;
    p=&Memory[allocp++];
    p->id=0;
    for(int i=0;i<26;i++)
    p->next[i]=NULL;
    return p;
}
void InsertTire(char *str,int ii)
{
    int i=0,k;
    TireNode *p=root;
    while(str[i])
    {
        k=str[i++]-'a';
        if(p->next[k]==NULL)
        p->next[k]=CreateTireNode();
        p=p->next[k];
    }
    p->id=ii;
}
int SearchTire(char *s)
{
    int i=0,k;
    TireNode *p=root;
    while(s[i])
    {
        k=s[i++]-'a';
        if(p->next[k]==NULL) return 0;
        p=p->next[k];
    }
    return p->id;
}
void init()
{
    root=CreateTireNode();
    for(int i=1;i<maxn;i++)
    fa[i]=i;
    memset(degree,0,sizeof(degree));
}
int Find_set(int x)
{
    if(x==fa[x]) return x;
    fa[x]=Find_set(fa[x]);
    return fa[x];
}
int main()
{
    char s1[15],s2[15];
    int cnt=1;
    init();
    while(scanf("%s %s",s1,s2)!=EOF)
    {
        int a=SearchTire(s1);
        if(!a)
        {
            InsertTire(s1,cnt);
            a=cnt++;
        }
        degree[a]++;
        int b=SearchTire(s2);
        if(!b)
        {
            InsertTire(s2,cnt);
            b=cnt++;
        }
        degree[b]++;
        int ra=Find_set(a);
        int rb=Find_set(b);
        if(ra!=rb) fa[rb]=ra;
    }
    if(cnt==0)
    printf("Possible\n");
    else
    {
        int c=0;
        for(int i=1;i<cnt;i++)
        if(i==fa[i]) c++;
        if(c>1)
        printf("Impossible\n");
        else
        {
            c=0;
            for(int i=1;i<cnt;i++)
            if(degree[i]&1) c++;
            if(c>2)
            printf("Impossible\n");
            else
            printf("Possible\n");
        }
    }
    return 0;
}