poj 2237(Catenyms)

 

这道题是，欧拉路和并查集结合起来的一道很恶心的题，最主要的难点在于，将这个模型抽象为欧拉路，已经应该用怎么样的存储方式。还有代码长了过后，变量的赋值一定要特别小心啊，我就是在一个地方把begin = i写成了begin - a； 于是就浪费了我好几个小时呢

Catenyms

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5806 Accepted: 1503

Description

A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms:

dog.gophergopher.ratrat.tigeraloha.alohaarachnid.dog

A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,

aloha.aloha.arachnid.dog.gopher.rat.tiger

Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.

Input

The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself.

Output

For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.

Sample Input

26alohaarachniddoggopherrattiger3oakmapleelm

Sample Output

aloha.arachnid.dog.gopher.rat.tiger***

Source

Waterloo local 2003.01.25

源代码：

#include<cstdio>#include<cstring>#include<string>#include<cstdio>#include<algorithm>#define MAXN  1010#define len 30using namespace std;char str[MAXN][len];char str1[MAXN][len];int father[len], n, changdu;struct orer{    int from, to, visit;}oulaa[MAXN];int cmp(const void *a, const void *b){    return strcmp((char *)a, (char *)b);}int find(int x){    int i,j;    i = x;    while(x != father[x])  x = father[x];    while(i != x){//完成状态压缩        j = father[i];        father[i] = x;        i = j;    }    return x;}void merge(int a, int b){    int x = find(a);    int y = find(b);    father[x] = y;}void eurlar(int start,int size){for(int i=1;i<=size;i++){if(!oulaa[i].visit && oulaa[i].from==start){oulaa[i].visit = 1;eurlar(oulaa[i].to, size);strcpy(str1[++changdu],str[i]);}}}int main(){    int t, in_out,a, b,begin, flag, in[len], out[len], used[len];    scanf("%d", &t);    while(t--){        //printf("t : %d\n", t);        memset(in,0,sizeof(in));        memset(out,0,sizeof(out));        memset(used, 0, sizeof(used));        scanf("%d", &n);        if(n<3){printf("***");printf("\n");continue;}        in_out = 0;        for(int i = 1; i <= n; i++){            scanf("%s", str[i]);        }        for(int i = 1; i <= 26; i++){            father[i] = i;        }        qsort(str+1, n, sizeof(str[0]), cmp);        for(int i = 1; i <= n; i++){            a = str[i][0] - 'a' + 1;            b = str[i][strlen(str[i]) - 1] - 'a' + 1;            oulaa[i].from = a;            oulaa[i].to = b;            oulaa[i].visit = 0;            in[b]++;            out[a]++;            merge(a,b);            used[a] = 1;            used[b] = 1;        }        /*        for(int  i = 1; i <= 26; i++){            if(in[i] != out[i])                in_out++;        }        if(in_out > 2){            printf("***\n");            continue;        }*/        a = 0;        b = 0;        for(int i = 1; i<= 26; i++){            if(father[i] == i && used[i])                b++;        }        if(b > 1){            printf("***\n");            continue;        }        a = 0,b = 0,flag = 0;        for(int i = 1; i <= 26; i++){            if(in[i] == out[i])  continue;            else if(in[i] - 1 == out[i])  a++;            else if(in[i] + 1 == out[i])  b++;            else {                flag = 1;                break;            }        }        if(flag == 1 || a > 1 || b > 1 || a != b){            printf("***\n");            continue;        }        begin = 1;        changdu = 0;        if(a == 1){            for(int i = 1; i <= n;i++){                if(out[oulaa[i].from] - in[oulaa[i].from] == 1){                    begin = i;                    break;                }            }        }        eurlar(oulaa[begin].from, n);        for(int i = changdu; i > 1; i--){            printf("%s.", str1[i]);        }        printf("%s\n", str1[1]);    }    return 0;}