poj2337

题意：给定一些单词，如果一个单词的尾字母与另一个的首字母相同则可以连接。问是否可以每个单词用一次，将所有单词连接，可以则输出字典序最小的序列。

分析：把每个字母作为一个结点。每个单词作为一条边。这样就可以把问题转化为欧拉路径。现判断欧拉路径是否存在。若存在则找到欧拉路径，找的方法是，先找到出度>入度（无欧拉回路）的结点，从起点开始，dfs（v,e），v是当前结点，e是到v的边。用vis数组记录某条边是否被访问过，从v开始继续搜索未访问过的边。每层dfs结束时把该层的边入栈。dfs整个结束后，把边依次出栈，得到的就是欧拉路径。

void dfs(int v, int e)
{
    vis[e] = true;
    for (int i = head[v]; i != -1; i = edge[i].next)
        if (!vis[i])
            dfs(edge[i].v, i);
    stk[top++] = e;
}

本题要求字典序最小，那么在dfs的时候就每次先访问字典序最小的边（原因很难说清），具体做法就是把边插入邻接表的时候冒泡一下，维护邻接表有序。当有欧拉回路时还要注意起点的选择，要使得路径的第一条边最小。

View Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 30
#define maxl 25
#define maxm 1005

struct Edge
{
    int v, next, id;
} edge[maxm];

int n, m;
char word[maxm][maxl];
int father[maxn];
int stk[maxm];
int in[maxn], out[maxn];
int head[maxn];
int ecount;
int top, s;
bool vis[maxm];
bool rep;

int getanc(int a)
{
    if (a == father[a])
        return a;
    return father[a] = getanc(father[a]);
}

void merge(int a, int b)
{
    father[getanc(a)] = getanc(b);
}

void addedge(int a, int b, int c)
{
    edge[ecount].next = head[a];
    edge[ecount].id = c;
    edge[ecount].v = b;
    head[a] = ecount++;
    int temp = head[a];
    while (edge[temp].next != -1)
    {
        int u = edge[temp].id;
        int v = edge[edge[temp].next].id;
        if (strcmp(word[u], word[v]) > 0)
        {
            swap(edge[temp].v, edge[edge[temp].next].v);
            swap(edge[temp].id, edge[edge[temp].next].id);
        }
        temp = edge[temp].next;
    }
}
void input()
{
    for (int i = 0; i < n; i++)
        father[i] = i;
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    ecount = 0;
    memset(head, -1, sizeof(head));
    scanf("%d", &m);
    for (int i = 0; i < m; i++)
    {
        scanf("%s", word[i]);
        int a = word[i][0] - 'a';
        int b = word[i][strlen(word[i]) - 1] - 'a';
        addedge(a, b, i);
        merge(a, b);
        in[b]++;
        out[a]++;
    }
}

bool ok()
{
    s = 0;
    int cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < n; i++)
    {
        if (abs(in[i] - out[i]) > 1)
            return false;
        if (in[i] - out[i] == 1)
            cnt1++;
        if (out[i] - in[i] == 1)
        {
            s = i;
            cnt2++;
        }
    }
    if (cnt1 == 0 && cnt2 == 0)
        rep = true;
    else
        rep = false;
    if (cnt1 > 1 || cnt2 > 1 || cnt1 != cnt2)
        return false;

    cnt1 = 0;
    for (int i = 0; i < n; i++)
        if (i == father[i] && (in[i] | out[i]))
            cnt1++;
    if (cnt1 > 1)
        return false;

    bool first = true;
    if (rep)
        for (int i = 0; i < n; i++)
            if (head[i] != -1 && (first || strcmp(word[edge[head[s]].id], word[edge[head[i]].id]) > 0))
            {
                first = false;
                s = i;
            }
    return true;
}

void dfs(int v, int e)
{
    vis[e] = true;
    for (int i = head[v]; i != -1; i = edge[i].next)
        if (!vis[i])
            dfs(edge[i].v, i);
    stk[top++] = e;
}

void work()
{
    memset(vis, 0, sizeof(vis));
    top = 0;
    dfs(s, ecount);
}

void print()
{
    top--;
    printf("%s", word[edge[stk[top - 1]].id]);
    top--;
    while (top > 0)
    {
        printf(".%s", word[edge[stk[top - 1]].id]);
        top--;
    }
    putchar('\n');
}

int main()
{
    //freopen("t.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        n = 26;
        input();
        if (!ok())
        {
            printf("***\n");
            continue;
        }
        else
            work();
        print();
    }
    return 0;
}