Catenyms UVA - 10441（有向图的欧拉回路）

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这是一道有向图的欧拉回路，首先得判断是否能构成欧拉路径，使用邻接表进行存储边的关系，判断每个点的入度与出度，最后还得利用个并查集，判断是否确实能否构成拓扑排序，最后输出欧拉路径即可.

#include<bits/stdc++.h>

using namespace std;

bool app[30];
char res[1005][26];
int in[30], out[30], deg[30], father[30], adj[30];
int n, ant, odd, beginn, endd;

struct edge
{
    bool vis;
    char sh[25];
    int y, next;
}a[1005];

bool cmp(edge a, edge b)
{
    return strcmp(a.sh, b.sh) > 0;
}

void init()
{
    int i;
    ant = odd = 0;
    memset(app, false, sizeof(app));
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(deg, 0, sizeof(deg));
    memset(adj, -1, sizeof(adj));
    for(i = 0; i < 1005; i++) a[i].vis = false;
    for(i = 0; i < 26; i++) father[i] = i;
}

int find(int x)
{
    if(x != father[x]){
        father[x] = find(father[x]);
    }
    return father[x];
}

void Union(int x, int y)
{
    int t1=find(x);
    int t2=find(y);
    if(t1!=t2){
        father[t1]=t2;
    }
    return ;
}

int judge()     //判断是否满足欧拉。0不满足，1欧拉回路，2欧拉路
{
    int i, k;
    for(i = 0; i < 26; i++){     //判断有向图欧拉
        if(!app[i]) continue;
        deg[i] = in[i] - out[i];
        if(abs(deg[i]) > 1) return 0;
        if(deg[i] > 0) beginn = i;   //起点
        if(deg[i] < 0) endd = i;     //终点
        if(deg[i]%2) odd++;
        if(odd > 2) return 0;
    }
    for(i = 0; i < 26; i++) if(app[i]) break;
    k = find(i);
    for(i = 0; i < 26; i++){   //判断连通性
        if(!app[i]) continue;
        if(k != find(i)) return 0;
    }
    if(odd == 0){    //有欧拉回路
        for(i = 0; i < 26; i++)
            if(app[i]) break;
        beginn = i;
        return 1;
    }
    return 2;
}

void dfs(int x, int id) //深搜寻找欧拉路径，很巧妙！！！
{
    int i;
	for(i = adj[x]; i != -1; i = a[i].next){
		if(!a[i].vis){
			a[i].vis = true;
			dfs(a[i].y, i);
		}
	}
	if(id != -1) strcpy(res[ant++], a[id].sh);  //最先进去的肯定是终点
}

int main()
{
    int i, x, y, fx, fy, t, len, tar;
    scanf("%d", &t);
    while(t--){
        init();
        scanf("%d", &n);
        for(i = 0; i < n; i++){
            scanf("%s", a[i].sh);
        }
        //题目要求是字典顺序，但是我是用前插链表，这时的顺序恰好会相反
        sort(a, a+n, cmp);  //所以排序时从大到小，这样刚刚会是字典顺序
        for(i = 0; i < n; i++){
            //构建出in与out，并构图，以及构建并查集
            len = strlen(a[i].sh);
            x = a[i].sh[0] - 'a';
            y = a[i].sh[len-1] - 'a';
            in[x]++;
            out[y]++;
            app[x] = true;
            app[y] = true;
            a[i].y = y;
            a[i].next = adj[x];
            adj[x] = i;
            fx = find(x);
            fy = find(y);
            if(fx != fy) Union(fx, fy);
        }
        tar = judge();
        if(tar == 0){
            printf("***\n");
            continue;
        }
        dfs(beginn, -1);
        printf("%s", res[ant-1]);
        for(i = ant-2; i >= 0; i--){
            printf(".%s", res[i]);
        }
        printf("\n");
    }
    return 0;
}