图论算法
问题1、图的邻接矩阵表示法
#define MAXSIZE 100 //定义最大顶点个数
typedef char VertexType; //顶点的类型
struct Graph
{
VertexType Vertex[MAXSIZE]; //顶点
int AdjMatrix[MAXSIZE][MAXSIZE];//两个顶点之间的权重
int vexnum; //实际顶点数
int arcnum; //实际的弧数
};
struct Info
{
VertexType start;
VertexType end;
int weight;
};
void AddVextex(Graph *g){
for (int i = 0; ivexnum; i++){
g->Vertex[i] = 'A' + i;
}
}
void CreateGraph(Graph *g){
g->vexnum = 5;//确定顶点个数
g->arcnum = 6; //确定弧数
//弧长初始化为1000
for (int i = 0; ivexnum; i++){
for (int j = 0; jvexnum; j++){
g->AdjMatrix[i][j] = 1000;
}
}
Info info[6] = { 'A', 'B', 3,
'B', 'C', 6, 'C', 'D', 9, 'A', 'D', 12, 'D', 'E', 23, 'D', 'A', 13
};
for (int i = 0; i<6; i++){
g->AdjMatrix[info[i].start - 'A'][info[i].end - 'A'] = info[i].weight;
}
}
void Display(Graph g){
for (int i = 0; ifor (int j = 0; jcout << g.AdjMatrix[i][j] << "\t";
}
cout << endl;
}
} 问题2、图的深度优先遍历DFS
void DFS(Graph g){
int *flag = new int[g.vexnum];//标记顶点是否被访问
memset(flag, 0, sizeof(int) * 5);
flag[0] = 1;//访问第一个顶点
cout << "0" << "->";
stack<int> s;
s.push(0); //栈保存第一个顶点，先进后出
while (!s.empty()){
int cur = s.top();//栈中当前的顶点
int i = 0;
for (; i < g.vexnum; i++){
if (g.AdjMatrix[cur][i] < 1000 && flag[i] == 0){
flag[i] = 1;
cout << i << "->";
s.push(i);
break;
}
}
if (i==g.vexnum)
s.pop();
}
cout << endl;
}问题3、宽度优先遍历BFS
void BFS(Graph g){
int *flag=new int[g.vexnum];
memset(flag, 0, sizeof(int) * 5);
flag[0]=1;
cout<<"0->";
queue<int> q;
q.push(0);
while(!q.empty()){
int cur=q.front();
int index=-1;
for(int i=0;iif(g.AdjMatrix[cur][i]<1000&&flag[i]==0){
flag[i]=1;
cout<"->";
q.push(i);
}
}
if(index==-1){
q.pop();
}
}
} 问题4、图的最小生成树：在一个联通网的所有生成树中所有边的代价之和的那颗生成树。带权的最小生成树要求权重最小。Prime算法（针对于无向图）
void PrimeTree(Graph g,int start){
int *path=new int[g.vexnum];
memset(path,0,sizeof(int)*g.vexnum);
for(int i=0;ibool *tree=new bool[g.vexnum];
memset(tree,0,sizeof(bool)*g.vexnum);
tree[start]=1; //遍历起点
int next=0;
cout<"->";
int mindis=0;
for(int i=1;i//遍历下一个结点
int min=INT_MAX;
for(int j=0;jif(!tree[j]&&min>path[j]){
min=path[j];
next=j;
}
}
//更新path
tree[next]=1;
cout<"->";
mindis+=min;
for(int k=0;kif(!tree[k]&&path[k]>g.AdjMatrix[next][k]){
path[k]=g.AdjMatrix[next][k];
}
}
}
cout<cout<<"最小生成树权值为"< //问题5、最短路径Dijkstra算法
//求起点start到任何一点的最短路径
void Dijkstra(Graph g,int start){
int path[MAXSIZE]; //保存到当前结点的上一个结点是谁
int flag[MAXSIZE]; //标记顶点是否被访问过
memset(flag,0,sizeof(flag)); //
int dist[MAXSIZE]; //dist保存的是start到当前位置的最短路径值
for（int end=0;endif(dist[end]<1000)
path[end]=start; //初始化保存到达当前结点的上一个结点
else
path[end]=-1;
}
dist[start]=0; //将start到start的路径设为0
flag[start]=1; //访问了第一个结点，将结点并入s
int next=start;
for(int i=1;iint min=INT_MAX;
for(int j=0;jif(!flag[j]&&min>dist[j]){
next=j;
min=dist[j];
}
}
flag[next]=1; //将next并入s
//更新最短路径信息
for(int k=0;kif(!flag[k]&&min<1000&&g.AdjMatrix[next][k]<1000&&g.AdjMatrix[next][k]+minmin;
path[k]=next;
}
}
}
//打印路径信息
for (int i = 0; i < g.vexnum; i++){
cout << start << "到" << i << "的距离为："<",最短路径为:";
for (int j = path[i]; j != -1; j = path[j])
cout << j << "->";
cout << endl;
}
}
最后附上完整的代码
#includevexnum; i++){
g->Vertex[i] = 'A' + i;
}
}
void CreateGraph(Graph *g){
g->vexnum = 5;//确定顶点个数
g->arcnum = 6; //确定弧数
//弧长初始化为1000
for (int i = 0; ivexnum; i++){
for (int j = 0; jvexnum; j++){
g->AdjMatrix[i][j] = 1000;
}
}
Info info[8] = { 'A', 'B', 6, 'A', 'E',12,
'B', 'C', 8, 'C', 'D', 6, 'A', 'D', 7, 'E', 'D', 5, 'E', 'C', 4, 'E', 'B',9
};
for (int i = 0; i<8; i++){
g->AdjMatrix[info[i].start - 'A'][info[i].end - 'A'] = info[i].weight;
g->AdjMatrix[info[i].end - 'A'][info[i].start - 'A'] = info[i].weight;
}
}
void Display(Graph g){
for (int i = 0; ifor (int j = 0; jcout << g.AdjMatrix[i][j] << "\t";
}
cout << endl;
}
}
int FindNext(Graph g, int cur){
for (int i = 1; iif (g.AdjMatrix[cur][i]<1000)
return i;
}
return -1;
}
void DFS(Graph g){
int *flag = new int[g.vexnum];//标记顶点是否被访问
memset(flag, 0, sizeof(int) * 5);
flag[0] = 1;//访问第一个顶点
cout << "0" << "->";
stack<int> s;
s.push(0); //栈保存第一个顶点，先进后出
while (!s.empty()){
int cur = s.top();//栈中当前的顶点
int i = 0;
for (; i < g.vexnum; i++){
if (g.AdjMatrix[cur][i] < 1000 && flag[i] == 0){
flag[i] = 1;
cout << i << "->";
s.push(i);
break;
}
}
if (i==g.vexnum)
s.pop();
}
cout << endl;
}
void BFS(Graph g){
int *flag = new int[g.vexnum];
memset(flag, 0, sizeof(int) * 5);
flag[0] = 1;
cout << "0->";
queue<int> q;
q.push(0);
while (!q.empty()){
int cur = q.front();
int index = -1;
for (int i = 0; iif (g.AdjMatrix[cur][i]<1000 && flag[i] == 0){
flag[i] = 1;
cout << i << "->";
q.push(i);
}
}
if (index == -1){
q.pop();
}
}
cout << endl;
}
void PrimeTree(Graph g,int start){
int *path = new int[g.vexnum];
memset(path, 0, sizeof(int)*g.vexnum);
for (int i = 0; ibool *tree = new bool[g.vexnum];
memset(tree, 0, sizeof(bool)*g.vexnum);
tree[start] = 1; //遍历起点
int next = 0;
cout << start << "->";
int mindis = 0;
for (int i = 1; i//遍历下一个结点
int min = INT_MAX;
for (int j = 0; jif (!tree[j] && min>path[j]){
min = path[j];
next = j;
}
}
//更新path
tree[next] = 1;
cout << next << "->";
mindis += min;
for (int k = 0; kif (!tree[k] && path[k]>g.AdjMatrix[next][k]){
path[k] = g.AdjMatrix[next][k];
}
}
}
cout << endl;
cout << "最小生成树权值为" << mindis << endl;
}
//求起点start到任何一点的最短路径
void Dijkstra(Graph g, int start){
int path[MAXSIZE]; //保存start到end的最短路径
int flag[MAXSIZE]; //标记顶点是否被访问过
memset(flag, 0, sizeof(flag)); //
int dist[MAXSIZE]; //dist保存的是start到当前位置的最短路径值
for(int end = 0; end < g.vexnum; end++){
dist[end] = g.AdjMatrix[start][end];
//记录经过的路径
if (dist[end] < 1000)
path[end] = start;
else
path[end] = -1;
}
dist[start] = 0; //将start到start的路径设为0
flag[start] = 1; //访问了第一个结点，将结点并入s
int next = start;
for (int i = 1; iint min = INT_MAX;
for (int j = 0; jif (!flag[j] && min>dist[j]){
next = j;
min = dist[j];
}
}
flag[next] = 1; //将next并入s
//更新最短路径信息
for (int k = 0; kif (!flag[k] && min<1000 && g.AdjMatrix[next][k]<1000 && (g.AdjMatrix[next][k] + min)//保存的是到达当前的结点的上一个结点，不断更新
}
}
}
//打印路径信息
for (int i = 0; i < g.vexnum; i++){
cout << start << "到" << i << "的距离为："<",最短路径为:";
for (int j = path[i]; j != -1; j = path[j])
cout << j << "->";
cout << endl;
}
}
int main(){
Graph g;
CreateGraph(&g);
Display(g);
DFS(g);
BFS(g);
//PrimeTree(g,0);
Dijkstra(g, 0);
system("pause");
return 0;
}