UVa 10596 - Morning Walk, 赤裸裸的欧拉回路

最新推荐文章于 2025-12-02 15:12:10 发布

最新推荐文章于 2025-12-02 15:12:10 发布 · 130 阅读

文章标签：

#php #数据结构与算法

本文深入分析了一个名为Kamal的个体在从Dinajpur搬迁到Chittagong后，如何通过每天早晨的不同路径漫步来避免重复路线，并探究了他是否有可能在一次行走中覆盖所有道路。通过使用欧拉回路的概念，文章提供了解决该问题的方法。

10596-Morning Walk

3791

37.43%

1073

89.84%

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=105&page=show_problem&problem=1537

题目：

Kamalis aMotashotaguy. He has got a new job inChittagong. So, he has moved toChittagongfromDinajpur. He was getting fatter inDinajpuras he had no work in his hand there. So, moving toChittagonghas turned to be a blessing for him. Every morning he takes a walk through the hilly roads of charming cityChittagong. He is enjoying this city very much. There are so many roads inChittagongand every morning he takes different paths for his walking. But while choosing a path he makes sure he does not visit a road twice not even in his way back home. An intersection point of a road is not considered as the part of the road. In a sunny morning, he was thinking about how it would be if he could visit all the roads of the city in a single walk. Your task is to helpKamalin determining whether it is possible for him or not.

题目大意：

Kamal每天早上都要从家里走到Chittagongand这个地方。从家里到Chittagongand有很多条路，他每天早上都要先选择好一条路线，这条路线从家里走到Chittagongand，在从Chittagongand走回家里。这条路线不能重复地经过同一条路。两个地点间可能会有多条路。比如多次出现了从A到B的路线, 那么表示每次出现的都是不同的路

分析与总结：

这题就是赤裸裸的欧拉回路。UVa 10054 - The Necklace和 UVa 10129 - Play on Words 这两篇讲得较详细。不再累赘

1. 欧拉回路+DFS

#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
int vis[210],N, M, G[210][210], inDegree[210], outDegree[210];

void dfs(int v)         //深度优先遍历  
{  
    vis[v]=true;  
    for(int i=0;i<N;i++)  
    {  
        if(!vis[i] && G[v][i])  
        {  
            dfs(i);  
        }  
    }  
}  

int main(){
#ifdef LOCAL
    freopen("input.txt","r",stdin);
#endif
    while(~scanf("%d %d",&N, &M)){
        memset(G, 0, sizeof(G));
        int a,b;
        memset(vis, 0, sizeof(vis));
        for(int i=0; i<M; ++i){
            scanf("%d %d",&a,&b);
            G[a][b] = G[b][a] = 1;
            ++vis[a]; 
            ++vis[b];
        }
        
        int cnt = 0;
        for(int i=0; i<N; ++i){
            if(vis[i]%2==1){
                ++cnt;
                break;
            }
        }
        memset(vis, 0, sizeof(vis));
        if(cnt || M<2) printf("Not Possible\n");
        else{
            dfs(0);
            bool flag = true;
            for(int i=0; i<N; ++i) {
                if(!vis[i]) flag = false;
            }
            if(flag)printf("Possible\n");
            else printf("Not Possible\n");
        }
    
    }
    return 0;
}

2，欧拉回路+并查集

#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
int vis[210],N, M, G[210][210], inDegree[210], outDegree[210], f[210];

void init()   //并查集判断是否是一个连通图  
{  
    int i;  
    for(i=0;i<N;i++)  
        f[i]=i;  
}  
int find(int x)  
{  
    int r=x;  
    while(f[r]!=r)  
    r=f[r];  
    f[x]=r;  
    return r;  
}  
void Union(int x,int y)  
{  
    int fx,fy;  
    fx=find(x);  
    fy=find(y);  
    if(fx!=fy)  
    f[fx]=fy;  
}  

int main(){
#ifdef LOCAL
    freopen("input.txt","r",stdin);
#endif
    while(~scanf("%d %d",&N, &M) && N){
        memset(G, 0, sizeof(G));
        int a,b;
        init();
        memset(vis, 0, sizeof(vis));
        for(int i=0; i<M; ++i){
            scanf("%d %d",&a,&b);
            ++G[a][b];
            ++G[b][a];
            ++vis[a];
            ++vis[b];
            Union(a,b);
        }
        
        int ans=0;
        for(int i=0; i<N; ++i) if(f[i]==i) ++ans;
        
        if(ans==1 && M>=2){
            int cnt = 0;
            for(int i=0; i<N; ++i){
                if(vis[i]%2==1){
                    ++cnt;
                    break;
                }
            }
            if(cnt) printf("Not Possible\n");
            else printf("Possible\n");
        }
        else
            printf("Not Possible\n"); 
    }
    return 0;
}