（图6-9）Strongly Connected Components--C语言

Write a program to find the strongly connected components in a digraph.

函数接口定义：

void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) );

where Graph is defined as the following:

typedef struct VNode *PtrToVNode;
struct VNode {
Vertex Vert;
PtrToVNode Next;
};
typedef struct GNode *Graph;
struct GNode {
int NumOfVertices;
int NumOfEdges;
PtrToVNode *Array;
};

Here void (*visit)(Vertex V) is a function parameter that is passed into StronglyConnectedComponents to handle (print with a certain format) each vertex that is visited. The function StronglyConnectedComponents is supposed to print a return after each component is found.

裁判测试程序样例：

#include <stdio.h>
#include <stdlib.h>

#define MaxVertices 10  /* maximum number of vertices */
typedef int Vertex;     /* vertices are numbered from 0 to MaxVertices-1 */
typedef struct VNode *PtrToVNode;
struct VNode {
    Vertex Vert;
    PtrToVNode Next;
};
typedef struct GNode *Graph;
struct GNode {
    int NumOfVertices;
    int NumOfEdges;
    PtrToVNode *Array;
};

Graph ReadG(); /* details omitted */

void PrintV( Vertex V )
{
   printf("%d ", V);
}

void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) );

int main()
{
    Graph G = ReadG();
    StronglyConnectedComponents( G, PrintV );
    return 0;
}

/* Your function will be put here */

输入样例：

4 5
0 1
1 2
2 0
3 1
3 2

输出样例：

3 
1 2 0 

Note: The output order does not matter. That is, a solution like

0 1 2
3

is also considered correct.

代码：

void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) )
{
    int mp[MaxVertices][MaxVertices] = {0};//邻接矩阵
    int num = G -> NumOfVertices;//顶点数
    int vis[MaxVertices] = {0};
    //将邻接表转换成邻接矩阵
    for(int i = 0;i < num;i ++) {
        PtrToVNode p = G -> Array[i];
        while(p) {
            mp[i][p -> Vert] = 1;//经过的顶点都标记一遍
            p = p -> Next;
        }
    }
    for(int k = 0;k < num;k ++) {
        for(int i = 0;i < num;i ++) {
            for(int j = 0;j < num;j ++) {
                if(mp[i][k] && mp[k][j]) 
                    mp[i][j] = 1;// 如果存在 i->k->j 的路径，则标记 i->j 可达
            }
        }
    }
    for(int i = 0;i < num;i++) 
    {
        if(vis[i]) continue;//如果顶点已经访问过，则跳过
        visit(i);//访问当前顶点
        vis[i] = 1;//标记当前顶点已访问
        for(int j = 0;j < num;j ++) {
            if(!vis[j] && mp[i][j] && mp[j][i]) {
                vis[j] = 1;//标记与当前顶点同属一个强连通分量的顶点已访问
                visit(j);//访问这些顶点
            }
        }
        putchar('\n');//输出一个换行，表示一个强连通分量结束
    }
}