1241: Graph Connectivity

Graph Connectivity的求解算法

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该博客介绍了一个关于图连通性的编程问题，目标是找出给定图中的最大连通子图数量。输入是一个包含节点连接信息的字符序列，输出是连通子图的个数。程序使用深度优先搜索（DFS）进行遍历，并通过一个示例输入和输出展示了其工作原理。

1241: Graph Connectivity

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	551	201	Standard

Consider a graph G formed from a large number of nodes connected by edges. G is said to be connected if a path can be found in 0 or more steps between any pair of nodes in G. For example, the graph below is not connected because there is no path from A to C.

This graph contains, however, a number of subgraphs that are connected, one for each of the following sets of nodes: {A}, {B}, {C}, {D}, {E}, {A,B}, {B,D}, {C,E}, {A,B,D}

A connected subgraph is maximal if there are no nodes and edges in the original graph that could be added to the subgraph and still leave it connected. There are two maximal connected subgraphs above, one associated with the nodes {A, B, D} and the other with the nodes {C, E}.

Write a program to determine the number of maximal connected subgraphs of a given graph.

Input and Output

The first line of input contains a single uppercase alphabetic character. This character represents the largest node name in the graph. Each successive line contains a pair of uppercase alphabetic characters denoting an edge in the graph. The sample input section contains a possible input set for the graph pictured above.

Input is terminated by a blank line.

Write in the output the number of maximal connected subgraphs.

Sample Input

E
AB
CE
DB
EC

Sample Output

This problem is used for contest: 119 151

#include<stdio.h>
#include<string.h>
int a[50][50];
int vis[50];
int sum;
void dfs(int step)
{
    int i;
    vis[step]=1;
    for(i=1;i<=sum;i++)
    {
        if(vis[i]==0&&a[step][i]==1)
        {
            dfs(i);
        }
    }
}
int main()
{
    char ch;
    int i,j;
    while(scanf("%c",&ch)==1)
    {
        getchar();
        memset(vis,0,sizeof(vis));
        memset(a,0,sizeof(a));
         sum=ch-'A'+1;
        char ch1,ch2;
        while(scanf("%c",&ch1)==1)
        {
            if(ch1=='/n') break;
            scanf("%c",&ch2);
            getchar();
            i=ch1-'A'+1;
            j=ch2-'A'+1;
            a[i][j]=a[j][i]=1;
        }
        int count=0;
        for(i=1;i<=sum;i++)
        {
            if(vis[i]==0)
            {
                count++;
                 dfs(i);
            }
        }
        printf("%d/n",count);
    }
    return 0;
}