UVA 10129 Play on Words

题目大意：判断首字母与末尾字母形成的若安路是否形成欧拉通路

解题思路：判断是否连通，然后判断度数是否为偶数（起点和终点入度和出度可以相差1）

#include <stdio.h>#include <iostream>#include <string.h>#include <stack>using namespace std;stack<int> sta[26];int m;int in[26], out[26];int flag[26];int num;int all;void dfs(int a) {    flag[a] = 0;    num++;    if(num == all)        return;    while(!sta[a].empty()) {        int b = sta[a].top();        if(flag[b] == 1) {            dfs(b);        }        sta[a].pop();    }    return;}int main() {    int n;    cin >> n;    while(n--) {        for(int i = 0; i < 26; i++)            while(!sta[i].empty())                sta[i].pop();        memset(in, 0, sizeof(in));        memset(out, 0, sizeof(out));        memset(flag, 0, sizeof(flag));        int x, y;        cin >> m;        getchar();        char ch[1000];        all= 0;        for(int i = 0; i < m; i++) {            gets(ch);            x = ch[0] - 'a';            y = ch[strlen(ch)-1] - 'a';            out[x]++;            in[y]++;            sta[x].push(y);            sta[y].push(x);            if(flag[x] == 0) {                all++;                flag[x] = 1;            }            if(flag[y] == 0) {                all++;                flag[y] = 1;            }        }        num = 0;        dfs(x);        int fir = 0;        int fin = 0;        if(num == all) {            for(int i = 0; i < 26; i++) {                if(fir > 1)                    break;                if(fin > 1)                    break;                if(in[i] == out[i])                    continue;                if(in[i] - out[i] == 1)                    fir++;                else if(out[i] - in[i] == 1)                    fin++;                else {                    num = -1;                    break;                }            }        }        if(num == all && fir - fin == 0)            printf("Ordering is possible.\n");        else            printf("The door cannot be opened.\n");    }    return 0;}