poj_2513

来源：互联网发布：龙江网络佳木斯分公司编辑：程序博客网时间：2024/06/03 20:15

源链接：http://poj.org/problem?id=2513

题目大意就是给你n串有颜色的棍子，两端都有颜色，问是否能将所有的棍子连成一条直线，使得相连处的颜色相同的。

很明显是个欧拉回路的问题，对于这道题目，允许存在入度为奇数的点，不过数量只能为2(即这两种颜色要在两端)，或者是0(即两端的颜色也相同，只是不相连而已)。

还有个工作，就是对输入的字符串进行编号，我们在这道题目中只能采取Trie树编号的方法，用map会超时。。。不过很奇怪。。我用了map的结果是wa，，找了好久都没找出来错误。。后来只能重新用字典树写了一遍。。囧。。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,t;#define Mod 1000000007#define N 110#define M 1000100int fa[M],num[M];struct Trie{Trie *next[27];int num;Trie(){num = 0;}};Trie *head;int id;void Init(){//初始化for(int i=0;i<M;i++)fa[i] = i;memset(num,0,sizeof num);head = new Trie();for(int i=0;i<26;i++)head->next[i] = 0;id = 1;}int Tree_Insert(char *s){//字典树编号Trie *tmp = head;Trie *t;int len = strlen(s);for(int i=0;i<len;i++){int h = s[i] - 'a';if(tmp->next[h] == NULL){t = new Trie;for(int i=0;i<26;i++)t->next[i] = NULL;tmp->next[h] = t;}tmp = tmp->next[h];}if(!tmp->num)tmp->num = id++;return tmp->num;}int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}void merge(int x,int y){int fx = find(x);int fy = find(y);if(fx!=fy){if(fx<fy)fa[fy] = fx;elsefa[fx] = fy;}}bool check(){int s = find(1);//判断是否只有一条直线,异或是有独立的for(int i=1;i<id;i++)if(fa[i] != s){return false;}int res = 0;for(int i=1;i<id;i++){//此题要求排成一条直线，那么入度是奇数的数量只能为2(即那两种颜色在两端)，或是0(即两端的颜色也相同，只是不相连而已)if(num[i]%2)res++;}return (res==0||res==2);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    Init();    char a[15],b[15];    while(scanf("%s%s",a,b)!=EOF){    int h1 = Tree_Insert(a);    int h2 = Tree_Insert(b);    num[h1]++;    num[h2]++;    merge(h1,h2);//欧拉回路    }    if(check())    printf("Possible\n");    else    printf("Impossible\n");return 0;}

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