国外大学的实验报告 formal reportpart3

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ALGORITHMIC APPROACH TO PROBLEM SOLVING
The result of output is shown as figure 9:
Figure 9.Output result for requirement 2 Requirement 3: The vertices of each connected component of a given graph. A vertex p is the parent of vertex c if c is visited from p, so the parent of vertex can be found before it goes to the its adjacent vertex. As we can see in sub-function “dfs(int)”, the parameter parent[t]=i; all the parent of current vertex has been stored in this array, but there is a special case that if the vertex is the first vertex in the component, it doesn’t have parent. So if else function will be used in main program: void main(void) { …..(previous program unchanged) else{ for(i=1; i<=component; i++){ printf("\ncomponnent %d is(vertex) ",i); for(k=start[i]; k<=end[i]; k++){ printf("%d ",sequence[k]); } } } // print out the component of the disconnected graph for requirement 2 for(i=1;i<=V;i++){ if(parent[i]==0){ printf("Vertex %d has no parent.\n",i); }//if the vertex is the first vertex in graph, no value will store into relevant parent[i] array, so it will be“0” by default.
ALGORITHMIC APPROACH TO PROBLEn called “dfs(int)”, it’s a recursive function used to search adjacent vertex of the given vertex which is not visited before, then set this adjacent vertex to be new start vertex and search the adjacent vertex recursively. The concept of follow chart for DFS subfunction is shown as below:
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ALGORITHMIC APPROACH TO PROBLEM SOLVING
The source code for dfs(int) sub-function: void dfs(int i){ int t; count=count+1; // Once visit the vertex, the count will increase by 1 to record how many vertex totally sequence[count]=i; //This sequence[count] array used to record the sequence of vertex visited visited[i]=1; //Once the vertex i is visited, it will mark as “1 ”, so it will not enter the loop even graph[i][t]=1 for(t=1; t<=V;t++){ if(a[i][t]==1 && visite[t]==0){ parent[t]=i; //Before go to the next vertex, record its parent to a array, it can used to print out the parent of each vertex for requirement 3 dfs(t); //recursively to find the next adjacent vertex } //Find the vertex which is adjacent to “i” and that vertex not visited before } } c) Construct over all program In order to implement the DFS, we have the construct a main program using the sub-routine we build before. For the main program, we need to print out the sequence of vertices which is related to array sequence[maxV], the parameter component will determine whether the graph is connected or not, if component=1, the graph is connected, otherwise, there are multiple disconnected components. The last requirement is to output the parent of vertex, in my program, I have set a array parent[maxV] to store the parent of value for each vertex, once the vertex is visited in “dfs(int)” sub-function.
Figure 8 .Output result for requirement 1
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ALGORITHMIC APPROACH TO PROBLEM SOLVING
Requirement 2: The vertices of each connected component of a given graph. The main program will be edited: void main(void) { …..(previous program unchanged) for(i=1;i<=V;i++){ if(visited[i]==0){ component = component+1; //Once come into the loop again that the disconnected component quantity will increase by 1. start[component]=count; //record the start index of nth component dfs(i); end[component]=count; //record the end index of nth component } } printf("\nvisit secquence is "); for(i=1; i<=V; i++) printf("%d ",sequence[i]); //print sequence if(component == 1){ printf("The graph is connected\n"); printf("Component is "); for(k=start[1]; k<=end[1]; k++){ printf("%d ",sequence[k]); } //component = 1,graph is connected, directly print out. } else{ for(i=1; i<=component; i++){ printf("\ncomponnent %d is(vertex) ",i); for(k=start[i]; k<=end[i]; k++){ printf("%d ",sequence[k]); } } } //Use start[maxV] and end[maxV] to print out the component of the disconnected graph } The component value can be used to indicate the graph is connected or not and how many components of graph have. When the program process run out from dfs(int) sub-function, that means one component graph formed, if there are any vertex not visited, the loop will go on running and find out the second component. So, for each component, I set start and end index to record which vertexes are belong which component graph. Finally, use two for loop to print out the component.