答案在風(fēng)中飄蕩抒情作文范文
1、OSI七層模型
2物理層3數據鏈路層4網(wǎng)絡(luò )層5傳輸層6會(huì )話(huà)層7表示層8應用層
2、進(jìn)程間數據共享的方式三種
文件映射
共享內存
信號
匿名管道
命名管道
郵件槽
剪貼板
3、TCP/UDP區別
4、打印數組的所有組合
[cpp] view plaincopyprint?
#include
int n = 0;
void swap(int *a, int *b)
{
int m;
m = *a;
*a = *b;
*b = m;
}
void perm(int list[], int k, int m)
{
int i;
if(k > m)
{
for(i = 0; i <= m; i++)
printf("%d ", list[i]);
printf(" ");
n++;
}
else
{
for(i = k; i <= m; i++)
{
swap(&list[k], &list[i]);
perm(list, k + 1, m);
swap(&list[k], &list[i]);
}
}
}
int main()
{
int list[] = {1, 2, 3, 4, 5};
perm(list, 0, 4);
printf("total:%d ", n);
return 0;
}
#include
int n = 0;
void swap(int *a, int *b)
{
int m;
m = *a;
*a = *b;
*b = m;
}
void perm(int list[], int k, int m)
{
int i;
if(k > m)
{
for(i = 0; i <= m; i++)
printf("%d ", list[i]);
printf(" ");
n++;
}
else
{
for(i = k; i <= m; i++)
{
swap(&list[k], &list[i]);
perm(list, k + 1, m);
swap(&list[k], &list[i]);
}
}
}
int main()
{
int list[] = {1, 2, 3, 4, 5};
perm(list, 0, 4);
printf("total:%d ", n);
return 0;
}
二進(jìn)制
首先，把數組每一個(gè)元素用一個(gè)二進(jìn)位表示，例如：
A B C D E
1 1 1 1 1 ---> 于是它最多有11111(二進(jìn)制)種不重復組合(即31種)(不考慮順序--按樓主要求)
于是，只要檢查從1到31這些數字的二進(jìn)位哪些是二進(jìn)制值1，就可以得出組合了。(位值為1的元素選取，位值為0的元素棄之)
(轉自網(wǎng)絡(luò ))
5、二叉樹(shù)的面積
深度優(yōu)先搜索，廣度優(yōu)先搜索的實(shí)現
[cpp] view plaincopyprint?
#include
#include
#include
#define TRUE 1
#define FLASE 0
#define OK 1
#define ERROR 0
#define INFEASIBLE -1
#define OVERFLOW -2
typedef int Status;
typedef int TElemType;
typedef struct BiTNode
{
TElemType data;
struct BiTNode *lchild,*rchild;
} BiTNode,*BiTree;
Status CreateBiTree(BiTree &T)
{
TElemType e;
scanf("%d",&e);
if(e==0) T=NULL;
else
{
T=(BiTree)malloc(sizeof(BiTNode));
if(!T)
exit(OVERFLOW);
T->data =e;
CreateBiTree(T->lchild );
CreateBiTree(T->rchild );
}
return OK;
}
int max(int a[])
{
int max,i;
max=a[0];
for(i=1;i<20;i++)
{
if(max
max=a[i];
}
return max;
}
int BiTreeWidth (BiTree T)
{
if(T==NULL)
return 0;
else
{
static int a[20]={0};
static int i=0;
a[i]++;
i++;
BiTreeWidth (T->lchild );
if(T->lchild ==NULL)
i--;
BiTreeWidth (T->rchild );
if(T->rchild ==NULL)
i--;
return max(a);
}
}
#include
#include
#include
#define TRUE 1
#define FLASE 0
#define OK 1
#define ERROR 0
#define INFEASIBLE -1
#define OVERFLOW -2
typedef int Status;
typedef int TElemType;
typedef struct BiTNode
{
TElemType data;
struct BiTNode *lchild,*rchild;
} BiTNode,*BiTree;
Status CreateBiTree(BiTree &T)
{
TElemType e;
scanf("%d",&e);
if(e==0) T=NULL;
else
{
T=(BiTree)malloc(sizeof(BiTNode));
if(!T)
exit(OVERFLOW);
T->data =e;
CreateBiTree(T->lchild );
CreateBiTree(T->rchild );
}
return OK;
}
int max(int a[])
{
int max,i;
max=a[0];
for(i=1;i<20;i++)
{
if(max
max=a[i];
}
return max;
}
int BiTreeWidth (BiTree T)
{
if(T==NULL)
return 0;
else
{
static int a[20]={0};
static int i=0;
a[i]++;
i++;
BiTreeWidth (T->lchild );
if(T->lchild ==NULL)
i--;
BiTreeWidth (T->rchild );
if(T->rchild ==NULL)
i--;
return max(a);
}
}
另附：
二叉樹(shù)高度、寬度、結點(diǎn)個(gè)數、葉子結點(diǎn)個(gè)數
實(shí)現二叉樹(shù)寬度遞歸算法~
#include
using namespace std;
typedef struct node
{
char data;
int lab;
struct node *lchild;
struct node *rchild;
}btree;
int m=0;
void ct(btree *&b,char *str)
{
btree *st[99],*p=NULL;
int top=-1,k,j=0;
char ch;
b=NULL;
ch=str[j];
while(ch!='\0')
{
switch(ch)
{
case '(':top++;st[top]=p;k=1;break;
case ')':top--;break;
case ',':k=2;break;
default:p=(btree *)malloc(sizeof(btree));
p->data=ch;
p->lchild=p->rchild=NULL;
if(b==NULL)
b=p;
else
{
switch(k)
{
case 1:st[top]->lchild=p;break;
case 2:st[top]->rchild=p;break;
}
}
}
j++;
ch=str[j];
}
}
void outbt(btree *b)
{
if(b!=NULL)
{
cout<
outbt(b->lchild);
outbt(b->rchild);
}
}
btree *findchild(btree *b,char x)
{
btree *p;
if(b==NULL)
{
return NULL;
}
else
if(b->data==x)
{
cout<<"找到結點(diǎn)"<
if(b->lchild==NULL)
cout<<"左節點(diǎn)為空!"<
else
cout<<"左孩子為:"<
if(b->rchild==NULL)
cout<<"右孩子為空!"<
else
cout<<"右孩子為:"<
return b;
}
else
{
p=findchild(b->lchild,x);
if(p!=NULL)
return p;
else
return findchild(b->rchild,x);
}
}
int btreeheight(btree *b)
{
int lchildh,rchildh;
if(b==NULL)
return(0);
else
{
lchildh=btreeheight(b->lchild);
rchildh=btreeheight(b->rchild);
return(lchildh>rchildh?(lchildh+1):(rchildh+1));
}
}
int i=-1,a[20];
void btreewide(btree *b)
{
if(b!=NULL)
{
if(b->lchild!=NULL)
{
i++;
b->lchild->lab=b->lab+1;
a[i]=b->lab+1;
}
if(b->rchild!=NULL)
{
i++;
b->rchild->lab=b->lab+1;
a[i]=b->lab+1;
}
btreewide(b->lchild);
btreewide(b->rchild);
}
}
void vernum(btree *b)
{
if(b!=NULL)
{
m++;
vernum(b->lchild);
vernum(b->rchild);
}
}
int leafver(btree *b)
{
if(b==NULL)
return 0;
else
if(b->lchild==NULL&&b->rchild==NULL)
return 1;
else
return leafver(b->lchild)+leafver(b->rchild);
}
void main()
{
char *s;
s="A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))";
btree *bt;
cout<<"將要創(chuàng )建的二叉樹(shù):"<
ct(bt,s);
cout<<"輸出二叉樹(shù):"<
outbt(bt);
cout<
cout<<"H結點(diǎn)左右孩子結點(diǎn)值:"<
findchild(bt,'H');
cout<<"二叉樹(shù)的深度:"<
vernum(bt);
cout<<"二叉樹(shù)結點(diǎn)個(gè)數:"<
cout<<"二叉樹(shù)葉子結點(diǎn)個(gè)數:"<
bt->lab=1;
btreewide(bt);
int j,k,num,max=0;
for(j=1;j<=i+1;j++)
{
num=0;
for(k=0;k<=i;k++)
if(a[k]==j)
num++;
if(max
{
max=num;
}
}
cout<<"二叉樹(shù)寬度為:"<
}
實(shí)現二叉樹(shù)寬度遞歸算法~
#include
using namespace std;
typedef struct node
{
char data;
int lab;
struct node *lchild;
struct node *rchild;
}btree;
int m=0;
void ct(btree *&b,char *str)
{
btree *st[99],*p=NULL;
int top=-1,k,j=0;
char ch;
b=NULL;
ch=str[j];
while(ch!='\0')
{
switch(ch)
{
case '(':top++;st[top]=p;k=1;break;
case ')':top--;break;
case ',':k=2;break;
default:p=(btree *)malloc(sizeof(btree));
p->data=ch;
p->lchild=p->rchild=NULL;
if(b==NULL)
b=p;
else
{
switch(k)
{
case 1:st[top]->lchild=p;break;
case 2:st[top]->rchild=p;break;
}
}
}
j++;
ch=str[j];
}
}
void outbt(btree *b)
{
if(b!=NULL)
{
cout<
outbt(b->lchild);
outbt(b->rchild);
}
}
btree *findchild(btree *b,char x)
{
btree *p;
if(b==NULL)
{
return NULL;
}
else
if(b->data==x)
{
cout<<"找到結點(diǎn)"<
if(b->lchild==NULL)
cout<<"左節點(diǎn)為空!"<
else
cout<<"左孩子為:"<
if(b->rchild==NULL)
cout<<"右孩子為空!"<
else
cout<<"右孩子為:"<
return b;
}
else
{
p=findchild(b->lchild,x);
if(p!=NULL)
return p;
else
return findchild(b->rchild,x);
}
}
int btreeheight(btree *b)
{
int lchildh,rchildh;
if(b==NULL)
return(0);
else
{
lchildh=btreeheight(b->lchild);
rchildh=btreeheight(b->rchild);
return(lchildh>rchildh?(lchildh+1):(rchildh+1));
}
}
int i=-1,a[20];
void btreewide(btree *b)
{
if(b!=NULL)
{
if(b->lchild!=NULL)
{
i++;
b->lchild->lab=b->lab+1;
a[i]=b->lab+1;
}
if(b->rchild!=NULL)
{
i++;
b->rchild->lab=b->lab+1;
a[i]=b->lab+1;
}
btreewide(b->lchild);
btreewide(b->rchild);
}
}
void vernum(btree *b)
{
if(b!=NULL)
{
m++;
vernum(b->lchild);
vernum(b->rchild);
}
}
int leafver(btree *b)
{
if(b==NULL)
return 0;
else
if(b->lchild==NULL&&b->rchild==NULL)
return 1;
else
return leafver(b->lchild)+leafver(b->rchild);
}
void main()
{
char *s;
s="A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))";
btree *bt;
cout<<"將要創(chuàng )建的二叉樹(shù):"<
ct(bt,s);
cout<<"輸出二叉樹(shù):"<
outbt(bt);
cout<
cout<<"H結點(diǎn)左右孩子結點(diǎn)值:"<
findchild(bt,'H');
cout<<"二叉樹(shù)的深度:"<
vernum(bt);
cout<<"二叉樹(shù)結點(diǎn)個(gè)數:"<
cout<<"二叉樹(shù)葉子結點(diǎn)個(gè)數:"<
bt->lab=1;
btreewide(bt);
int j,k,num,max=0;
for(j=1;j<=i+1;j++)
{
num=0;
for(k=0;k<=i;k++)
if(a[k]==j)
num++;
if(max
{
max=num;
}
}
cout<<"二叉樹(shù)寬度為:"<
}
6、地圖的點(diǎn)的文字位置的確定
1、掃描點(diǎn)附近的矩形鄰域內是否為空
2改進(jìn)：搜索算法
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