POJ 1386 Play on Words

欧拉回路

给出n个单词，如果一个单词的首字母和另一个单词的尾字母相同，则可以相连成一条边。用欧拉的思想做，把一个单词的首字母和尾字母作为一个顶点，如果满足条件，就连出一条有向边，记录出度和入度，用并查集来判断是否为一个强连通分支，如果是，再判断奇数度是否为2个或者0个。

AC代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define max_len 1010
#define maxv 30

using namespace std;

int in[maxv]; 
int out[maxv]; 
int p[maxv];
bool used[maxv];

void init() {
    for(int i = 0; i < maxv; ++i) {
        in[i] = 0;
        out[i] = 0;
        p[i] = -1;
        used[i] = 0;
    }
}
int find_set(int x) {
    if(0 <= p[x]) {
        p[x] = find_set(p[x]);
        return p[x];
    }
    return x;
}
void union_set(int a, int b) {
    int r1 = find_set(a);
    int r2 = find_set(b);
    if(r1 == r2)
        return;
    int n1 = p[r1];
    int n2 = p[r2];
    if(n1 < n2) {
        p[r2] = r1;
        p[r1] += n2;
    }
    else {
        p[r1] = r2;
        p[r2] += n1;
    }
}
int main() {
    int t = 0;
    char word[max_len];
    scanf("%d", &t);
    while(t--)
    {
	int n, i;
        init();
        scanf("%d", &n);
        for(i = 0; i < n; ++i)
        {
            scanf("%s", word);
            int len = strlen(word);
            int s = word[0] - 'a';
            int e = word[len - 1] - 'a';
            used[s] = 1;
            used[e] = 1;
            out[s]++;
            in[e]++;
            union_set(s, e);
        }
       
        int scc = 0;
        for(i = 0; i < maxv; ++i) {
            if(used[i] && 0 > p[i])
                ++scc;
        }
        if(1 < scc) {
            printf("The door cannot be opened.\n");
            continue;
        }
        int a = 0;
        int b = 0;
        for(i = 0; i < maxv; ++i) {
            if(used[i] && in[i] != out[i]) {
                if(1 == (in[i] - out[i]))
                    ++a;
                else if(1 == (out[i] - in[i]))
                    ++b;
                else
                    break;
            }
        }
        if(i < maxv)
            printf("The door cannot be opened.\n");
        else if(0 == (a + b) || (1 == a && 1 == b))
            printf("Ordering is possible.\n");
        else 
            printf("The door cannot be opened.\n");
    }
    return 0;
}