Colored Sticks

题意：就是求一个欧拉回路。
题解：本题是判断欧拉通路是否存在，但是如果是用map的话就会超时，这里采用了trie树，有发现trie树的一种用法。神奇啊
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define maxn 1100000
  6 using namespace std;
  7 const int N=3e5+100;
  8 int head[N],cnt,num,vis[N],deg[N];
  9 struct Node{
 10   int v;
 11   int next;
 12 }edge[N*3];
 13 void init(){
 14   memset(head,-1,sizeof(head));
 15   cnt=num=0;
 16   memset(vis,0,sizeof(vis));
 17   memset(deg,0,sizeof(deg));
 18 }
 19 
 20 void add(int u,int v){
 21     edge[cnt].v=v;
 22     edge[cnt].next=head[u];
 23     head[u]=cnt++;
 24     edge[cnt].v=u;
 25     edge[cnt].next=head[v];
 26     head[v]=cnt++;
 27 }
 28 void DFS(int u){
 29     if(vis[u])return;
 30     vis[u]=1;
 31     for(int i=head[u];i!=-1;i=edge[i].next){
 32         DFS(edge[i].v);
 33     }
 34 }
 35 struct Nod {        //0为无效值
 36     int lnk[26], val;
 37     void init() {
 38         memset(lnk, 0, sizeof(lnk));
 39         val = 0;
 40     }
 41 };
 42 const char BASE = 'a';
 43 struct Trie {
 44     Nod buf[maxn];
 45     int len;
 46     void init() {
 47         buf[len=0].init();
 48     }
 49     int insert(char * str, int val) {
 50         int now = 0;
 51         for(int i = 0; str[i]; i ++) {
 52             int & nxt = buf[now].lnk[str[i]-BASE];
 53             if(!nxt)    buf[nxt=++len].init();
 54             now = nxt;
 55         }
 56         buf[now].val = val;
 57         return now;
 58     }
 59     int search(char * str) {
 60         int now = 0;
 61         for(int i = 0; str[i]; i ++) {
 62             int & nxt = buf[now].lnk[str[i]-BASE];
 63             if(!nxt)    return 0;
 64             now = nxt;
 65         }
 66         return buf[now].val;
 67     }
 68 } trie;
 69 char s1[100],s2[100];
 70 int main(){
 71    trie.init();
 72    init();
 73    while(~scanf("%s%s",&s1,&s2)){
 74       if(trie.search(s1)==0)trie.insert(s1,++num);
 75       if(trie.search(s2)==0)trie.insert(s2,++num);
 76       int c1=trie.search(s1),c2=trie.search(s2);
 77       add(c1,c2);
 78       deg[c1]++;
 79       deg[c2]++;
 80    }
 81    bool flag=true;
 82    int ct=0;
 83     DFS(1);
 84     for(int i=1;i<=num;i++){
 85         if(!vis[i]){
 86             flag=false;
 87             break;
 88         }
 89     }
 90     for(int i=1;i<=num;i++){
 91         if(deg[i]%2==1)
 92             ct++;
 93     }
 94     if(flag){
 95         if(ct==0||ct==2){
 96             printf("Possible
");
 97         }
 98         else
 99             printf("Impossible
");
100     }
101     else
102             printf("Impossible
");
103 
104 }
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