POJ 2513 Colored Sticks(字典树+欧拉回路)

Colored Sticks

Time Limit: 5000MS Memory Limit: 128000KTotal Submissions: 37665 Accepted: 9873

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 redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

Hint

Huge input,scanf is recommended.

Source

The UofA Local 2000.10.14

题目大意：

给你好多木棍，两端涂上颜色，如果两端颜色相同就可以接在一起。问是否可以接成一个长木棍。

分析：

其实是字典树+欧拉回路判断的水题。自己纠结了半天的骚搞……最后WA了很多发。

大概就是字典树建树然后用字典树hash建图， 如果有颜色ab的木棍， 就把a和b连上边，最后判断图是否可以一笔画遍历所有的边即可。记得判断空集，空集需要输出可以。

（据说数据范围有问题，因为WA的次数太多也不知道是不是有问题了)

#include <iostream>#include <map>#include <queue>#include <algorithm>#include <cstdio>#include <cstring>#define MAX 600500using std::queue;struct node{int num, nxt[26];}tire[MAX * 27]; int top, htop;int nxt[MAX];bool  flg[MAX];int get(char str[]){int len = strlen(str), i, now = 0;for(i = 0; i < len; i++){if(tire[now].nxt[str[i] - 'a'] == 0){tire[now].nxt[str[i] - 'a'] = top++;}now = tire[now].nxt[str[i] - 'a'];}if(tire[now].num == 0) tire[now].num = htop++;return tire[now].num;}struct node2{int t, nxt;}edge[MAX * 2]; int head[MAX], inde, deg[MAX], oc[MAX];void fresh(){memset(head, -1, sizeof(head));inde = 0;}void addedge(int s,int t){edge[inde].t = t;edge[inde].nxt = head[s];head[s] = inde++;deg[s]++;}bool judge(){queue<int> que;int i, now, cnt;que.push(1);oc[1] = 1;cnt = 1;while(!que.empty()){now = que.front();que.pop();i = head[now];while(i >= 0){if(!oc[edge[i].t]){oc[edge[i].t] = 1;que.push(edge[i].t);cnt++;}i = edge[i].nxt;}}if(cnt != htop - 1) return false;cnt = 0;for(i = 1; i < htop; i++){if(deg[i] & 1) cnt++;}if(cnt > 2) return false;return true;}int main(){top = htop = 1;char s1[11], s2[11];int h1, h2, i, cnt;fresh();while(~scanf("%s%s", s1, s2)){h1 = get(s1);h2 = get(s2);addedge(h1, h2);addedge(h2, h1);}if(judge() || htop == 1)printf("Possible\n");else printf("Impossible\n");return 0;}