POJ_2513_Colored Sticks（欧拉路+字典树）
Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
| Total Submissions: 31721 | Accepted: 8391 |
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
题意：给你一些个木棒，每根木棒两端都着了 色，问能否将它们排成一条线，使得接触点的颜色是一样的。
解析：很明显的求欧拉路问题。一开始用map转化超时，后来去学了用字典树的方法处理。值得注意的是，trie数组得开足够大，我是开550000。反正在这个数组上面RE了n次心都碎了。
题目链接：http://poj.org/problem?id=2513
代码清单：
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
struct edge{
int point;
int letter[30];
}trie[550000];
int number=0,Point=0;
int father[550000];
char s1[15],s2[15];
int degree[550000];
int Find(int x){
if(x!=father[x]) return Find(father[x]);
return father[x];
}
int GetPoint(char s[]){
int j=0;
int len=strlen(s);
for(int i=0;i<len;i++){
if(trie[j].letter[s[i]-'a'])
j=trie[j].letter[s[i]-'a'];
else{
trie[j].letter[s[i]-'a']=++number;
j=number;
}
}
if(trie[j].point==0) trie[j].point=++Point;
return trie[j].point;
}
bool IsEuler(int n){
int cnt=0,temp=0;
for(int i=1;i<=n;i++){
if(father[i]==i) cnt++;
if(degree[i]%2) temp++;
}
if(cnt>1) return false;
else{
if(temp==0||temp==2) return true;
else return false;
}
}
int main(){
memset(trie,0,sizeof(trie));
memset(degree,0,sizeof(degree));
for(int i=0;i<550000;i++) father[i]=i;
//freopen("liuchu.txt","r",stdin);
while(scanf("%s%s",s1,s2)!=EOF){
int point1=GetPoint(s1);
int point2=GetPoint(s2);
degree[point1]++;
degree[point2]++;
point1=Find(point1);
point2=Find(point2);
father[point1]=point2;
}
if(IsEuler(Point)) printf("Possible\n");
else printf("Impossible\n");
return 0;
字典树部分可以用指针来写，用数组的弊端就是容易RE。
代码清单：
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAX = 26;
struct trie{
int point;
trie *next[MAX];
};
trie *root=new trie;
int father[550000];
char s1[15],s2[15];
int degree[550000],Point=0;
int Find(int x){
if(x!=father[x]) return Find(father[x]);
return father[x];
}
int GetPoint(char s[]){
trie *p=root, *q;
int len=strlen(s), pos;
for(int i=0;i<len;i++){
pos=s[i]-'a';
if(p->next[pos]==NULL){
q=new trie;
q->point=0;
for(int j=0;j<MAX;j++)
q->next[j]=NULL;
p->next[pos]=q;
p=p->next[pos];
}
else{
p=p->next[pos];
}
}
if(p->point==0) p->point=++Point;
return p->point;
}
bool IsEuler(int n){
int cnt=0,temp=0;
for(int i=1;i<=n;i++){
if(father[i]==i) cnt++;
if(degree[i]%2) temp++;
}
if(cnt>1) return false;
else{
if(temp==0||temp==2) return true;
else return false;
}
}
int main(){
memset(degree,0,sizeof(degree));
for(int i=0;i<MAX;i++) root->next[i]=NULL;
for(int i=0;i<550000;i++) father[i]=i;
// freopen("liuchu.txt","r",stdin);
while(scanf("%s%s",s1,s2)!=EOF){
int point1=GetPoint(s1);
int point2=GetPoint(s2);
degree[point1]++;
degree[point2]++;
point1=Find(point1);
point2=Find(point2);
father[point1]=point2;
}
if(IsEuler(Point)) printf("Possible\n");
else printf("Impossible\n");
return 0;
}