HDU 1116 Play on Words



http://acm.hdu.edu.cn/showproblem.php?pid=1116

有向图寻找欧拉路径：

1.连通图

2.所有的点出度入度相等或有且只有一个点出度比入度大1（出发点），一个点入度比出度大1（终止点）。

有向图寻找欧拉回路：

1.连通图

2.所有的点出度等于入度。

本题是词语接龙的模型，等价于在有向图中判断是否存在欧拉路径。使用并查集判断图是否连通。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;char str[1010];int father[30],vis[30],n,in[30],out[30];void init(){    for(int i=0; i<30; i++){        father[i] = i;        vis[i] = 0;        out[i] = 0;        in[i] = 0;    }}int find(int x){    if(x == father[x])  return x;    return find(father[x]);}int main(){//    freopen("in.txt", "r", stdin);    int cas;    scanf("%d",&cas);    while(cas--){        init();        scanf("%d",&n);        for(int i=0; i<n; i++){            scanf("%s",str);            int x = str[0] - 'a',y = str[strlen(str) - 1] - 'a';            father[x] = father[y] = find(x);            vis[x] = vis[y] = 1;            out[x]++,in[y]++;        }        int r = 0;        for(int i=0; i<30; i++)            if(vis[i] && i == father[i])    r++;        if(r > 1){cout << "The door cannot be opened." << endl; continue;}        int x = 0, y = 0, z = 0, flag = 1;        for(int i=0; i<30; i++){            if(vis[i] && in[i] != out[i]){                if(in[i] - out[i] == 1) x++;                else if(out[i] - in[i] == 1)    y++;                else z++;            }        }        if(z)   cout << "The door cannot be opened." << endl;        else if((!x && !y) || (x == 1 && y == 1))   cout << "Ordering is possible." << endl;        else    cout << "The door cannot be opened." << endl;    }    return 0;}