HDU 1878 欧拉回路（判断欧拉回路）

来源：互联网发布：java技术总监面试题编辑：程序博客网时间：2024/04/28 18:56

题意：略

解题思路：判断是否连通，有没有度为奇数的点。并查集， DFS都可以，用链式向前星建图会超时，不知道为什么，邻接矩阵反而不超时。

欧拉回路

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10725 Accepted Submission(s): 3902

Problem Description

欧拉回路是指不令笔离开纸面，可画过图中每条边仅一次，且可以回到起点的一条回路。现给定一个图，问是否存在欧拉回路？

Input

测试输入包含若干测试用例。每个测试用例的第1行给出两个正整数，分别是节点数N ( 1 < N < 1000 )和边数M；随后的M行对应M条边，每行给出一对正整数，分别是该条边直接连通的两个节点的编号（节点从1到N编号）。当N为0时输入结
束。

Output

每个测试用例的输出占一行，若欧拉回路存在则输出1，否则输出0。

Sample Input

3 31 21 32 33 21 22 30

Sample Output

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 1005const int INF = 0x3f3f3f3f;int G[MAX_N][MAX_N];bool vis[MAX_N];int degree[MAX_N];int n, m;void dfs(int u){    vis[u] = true;    for(int i = 1 ; i <= n ; i++)    {        if(G[u][i] && !vis[i])        {            dfs(i);        }    }    return;}inline void init(){    memset(vis, false, sizeof(vis));    memset(G, 0, sizeof(G));    memset(degree, 0, sizeof(degree));}int main(){//    int n, m;    while(scanf("%d", &n) && n)    {        init();        scanf("%d", &m);        int u, v;        for(int i = 0 ; i < m ; i++)        {            u = read();            v = read();            G[u][v] = G[v][u] = 1;            degree[u]++;            degree[v]++;        }        dfs(1);        int flag = 0;        for(int i = 1 ; i <= n ;i++)        {            if(degree[i] % 2)            {                flag = 1;                break;            }            if(!vis[i])            {                flag = 1;                break;            }        }        if(flag)        {            printf("0\n");//            continue;        }        else        {            printf("1\n");        }    }    return 0;}

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 1005const int INF = 0x3f3f3f3f;int parent[MAX_N];int degree[MAX_N];inline void init(int n){    memset(degree, 0, sizeof(degree));    for(int i = 1 ; i <=  n + 1 ; i++)    {        parent[i] = i;    }}int find(int a){    if(parent[a] != a)    {        parent[a] = find(parent[a]);    }    return parent[a];}void unite(int a, int b){    int fa = find(a);    int fb = find(b);    if(fa == fb) return;    parent[fb] = fa;}int same(int a, int b){    return find(a) == find(b);}int main(){    int n, m;    while(scanf("%d", &n) && n)    {        init(n);        scanf("%d", &m);        for(int i = 0 ; i < m ;i++)        {            int u = read();            int v = read();            unite(u, v);            degree[u]++;            degree[v]++;        }        int flag = 0;        for(int i = 1 ;i <= n ;i++)        {            if(degree[i] % 2)            {                flag = 1;                break;            }        }        if(flag)        {            printf("0\n");            continue;        }        int tmp = parent[1];        for(int i = 2 ; i <= n ; i++)        {            if(tmp != find(i))            {                flag = 1;                break;            }        }        if(flag)        {            printf("0\n");            continue;        }        printf("1\n");    }    return 0;}

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