CQUPT 的某数据结构homework
- 基于线性表的图书信息管理
- 基于栈的算术表达式求值
- 基于字符串模式匹配算法的病毒感染检测问题
- 基于哈夫曼树的数据压缩算法
- 基于二叉树的表达式求值算法
- 基于 Dijsktra 算法的最短路
- 基于广度优先搜索的六度空间
- 排序算法的实现与分析
基于线性表的图书信息管理
首先，因为我们要输入输出汉语，所以需要更改编码成为utf-8
我使用的是devcpp
所以其他版本编译器更改编码的就请自己去查询更改方式。
工具：
编译选项：
编译时加入以下命令，打勾，下面粘上此代码
-std=c++11 -finput-charset=utf-8
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
struct node_book
{
string name_book;
string id_book;
float price_book;
//int myid;
};
unordered_map<string ,int>mmp;//去重用
list<node_book> books;
vector<node_book> high_books;
double evrage;
int books_num,idx;
void output(){
for(auto j:books){
cout<<j.id_book<<" "<<j.name_book<<" ";
printf("%.2f\n",j.price_book);
}
}
void add(string id,float price, string name){
node_book new_book;
new_book.id_book = id;
new_book.name_book = name;
new_book.price_book = price;
books.push_back(new_book);
books_num++;
evrage+= price;
}
void init(){
books_num = 0;
float price;
string name,id;
while(1){
cin>>id>>name>>price;
if(id=="0"&&name=="0"&&price==0)break;
add(id,price,name);
mmp[id]++;
// evrage += price;
}
}
void func1(){
init();
cout<<books_num<<endl;
output();
}
void func2(){
init();
evrage /= books_num;
for(auto& j:books){
float te = j.price_book;
if(te < evrage) j.price_book*=1.2;
else j.price_book*=1.1;
}
printf("%.2f\n",evrage);
cout<<books_num<<endl;
output();
}
void func3(){
init();
high_books.clear();
high_books.push_back(*books.begin());
float high_price = high_books[0].price_book;
for(auto j:books){
if(j.price_book>high_price){
high_books.clear();
high_books.push_back(j);
high_price = j.price_book;
}
else if(high_price==j.price_book){
high_books.push_back(j);
}
}
cout<<high_books.size()<<endl;
for(auto j: high_books){
cout<<j.id_book<<" "<<j.name_book<<" ";
printf("%.02f\n",j.price_book);
}
}
void func4(){
init();
float price;
string name,id;
int place,idx = 1;
cin>>place;
cin>>id>>name>>price;
node_book new_book;
new_book.id_book = id;
new_book.name_book = name;
new_book.price_book = price;
list<node_book>::iterator pos ;
pos = books.begin();
for(auto j:books){
idx++;
pos++;
if(idx==place){
books.insert(pos,new_book);
books_num++;
cout<<books_num<<endl;
output();
return;
}
}
cout<<"抱歉，入库位置非法!\n";
}
void func5(){
init();
float price;
string name,id;
int place,idx = 1;
cin>>place;
list<node_book>::iterator pos ;
pos = books.begin();
for(auto j:books){
idx++;
pos++;
if(idx==place){
books.erase(pos);
books_num--;
}
}
cout<<books_num<<endl;
output();
}
void func6(){
init();
int place;
list<node_book>::iterator pos ;
vector<list<node_book>::iterator> dela;
pos = books.begin();
for(auto j:books){
//cout<<mmp[j.id_book]<<endl;
if(mmp[j.id_book]>1){
mmp[j.id_book]--;
dela.push_back(pos);
books_num--;
}
pos++;
}
for(auto it:dela){
books.erase(it);
}
cout<<books_num<<endl;
output();
}
void tishi(){
printf("\n输入1测试：基于顺序（链式）存储结构的图书信息表的创建和输出\n");
cout<<"输入2测试：基于顺序（链式）存储结构的图书信息表的修改\n";
cout<<"输入3测试：基于顺序（链式）存储结构的图书信息表的最贵图书的查找\n";
cout<<"输入4测试：基于顺序（链式）存储结构的图书信息表的新图书的入库\n";
cout<<"输入5测试：基于顺序（链式）存储结构的图书信息表的旧图书的出库\n";
cout<<"输入6测试：基于顺序（链式）存储结构的图书信息表的图书去重\n";
cout<<"输入-1，退出程序\n";
}
int main(){
int op =1;
while(op!=0){
tishi();
cin>>op;
if(op== 1)
func1();
else if(op==2)
func2();
else if(op==3)
func3();
else if(op==4)
func4();
else if(op==5)
func5();
else if(op==6)
func6();
books.clear();
}
return 0;
}
基于栈的算术表达式求值
普通应用即可
#include<bits/stdc++.h>
using namespace std;
#define MAXSIZE 100
#define N 10
string S;
//运算符栈
unordered_map<char,int>mmp;
void init0(){
mmp['('] = mmp[')']=3;
mmp['*'] = mmp['/'] = 2;
mmp['+'] = mmp['-'] = 1;
}
stack<char> sta_op;
stack<int> sta_num;
int verify(char ch)
{
//cout<<ch<<endl;
if(ch=='+')
return 0;
else if(ch=='-')
return 1;
else if(ch=='*')
return 2;
else if(ch=='/')
return 3;
else if(ch=='(')
return 4;
else if(ch==')')
return 5;
else
{
return 100;
}
}
void push_num(int& idx){
int j = idx;
int temp = 0;
while(S[j]>='0'&&S[j]<='9'){
temp*=10;
temp+=S[j]-'0';
j++;
}
sta_num.push(temp);
idx = j-1;
}
int judge(int a,char e,int b)
{
//cout<<e<<endl;
if(e=='+')
return a+b;
else if(e=='-')
return a-b;
else if(e=='*')
return a*b;
else if(e=='/')
return a/b;
else
{
cout<<"error!"<<endl;
return -1;
}
}
void func1(int& idx){
char ch = S[idx];
if(ch=='('||ch=='+'||ch=='-')
sta_op.push(ch);//of course +-不需要考虑运算顺序
else if(ch == '*'||ch=='/'){
if(S[idx+1]>='0'&&S[idx+1]<='9'){
//*/在后面跟上的是数字的情况下，一定是可以直接进行运算的
//那么不妨在这种情况下面直接开算
int j = idx+1;
int temp = 0;
while(S[j]>='0'&&S[j]<='9'){
temp*=10;
temp+=S[j]-'0';
j++;
}
sta_num.push(temp);
idx = j-1;
//
int la = sta_num.top();
sta_num.pop();
int pre = sta_num.top();
sta_num.pop();
int ans = judge(pre,ch,la);
sta_num.push(ans);
}
else{
sta_op.push(ch);//只能等后面在来运算辣
//就是等下一个左括号运算完成并出栈的时候来运算此符号
}
}else if(ch==')'){
while(sta_op.top()!='('){
int la = sta_num.top();
sta_num.pop();
int pre = sta_num.top();
sta_num.pop();
int ans = judge(pre,sta_op.top(),la);
sta_op.pop();
sta_num.push(ans);
}
sta_op.pop();
if(sta_op.top()=='*'||sta_op.top()=='/'){//
//这个时候就要运算89行处的*/法辣
int la = sta_num.top();
sta_num.pop();
int pre = sta_num.top();
sta_num.pop();
int ans = judge(pre,sta_op.top(),la);
sta_op.pop();
sta_num.push(ans);
}
}
}
int func(){
cin>>S;
if(S=="=")return 0;
for(int i=0;S[i]!='\0';i++)
{
if(S[i]>='0'&&S[i]<='9')
push_num(i);
else
func1(i);
}
//*/法已经算完辣，+-法没有顺序可言，但是为了减少代码量就不优化辣。
while(!sta_op.empty()){
int la = sta_num.top();
sta_num.pop();
int pre = sta_num.top();
sta_num.pop();
int ans = judge(pre,sta_op.top(),la);
sta_op.pop();
sta_num.push(ans);
}
cout<<sta_num.top()<<endl;
sta_num.pop();
return 1;
}
int main()
{
int ti=1;
while(ti){
cout<<"\n请输入中缀表达式的算数表达式\n仅输入一个 = 时结束程序\n";
if(!func())break;
}
return 0;
}
基于字符串模式匹配算法的病毒感染检测问题
kmp即可
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 10, M = 1e6 + 10;
char p[N], s[M];
int n, m;
int ne[N];
void tishi(){
cout<<"请输入病毒的ONA序列和人的NDNA序列,用空格隔开\n";
cout<<"当输入的两个DNA序列为 0 0 时退出程序！\n";
}
bool solve(){
string x,y;
cin>>x>>y;
n = y.size();
m = x.size();
if0(n){
s[i+1]= y[i];
}
if0(m) p[i+1] = x[i];
if(x=="0"&&y=="0")return 0;
memset(ne,0,sizeof(ne));
for (int i = 2, j = 0; i <= m; i ++ ){
while (j && p[i] != p[j + 1]) j = ne[j];
if (p[i] == p[j + 1]) j ++ ;
ne[i] = j;
}
// 匹配
for (int i = 0, j = -1; i <= n; i ++ ){
while (j && s[i] != p[j + 1]) j = ne[j];
if (s[i] == p[j + 1]) j ++ ;
if(j==m){
cout<<"YES\n";
return 1;
}
if (j == m)
{
j = ne[j];
// 匹配成功后的逻辑
}
}
cout<<"NO\n";
return 1;
}
int main(){
int t=1;
while (t)
{
tishi();
if(!solve()) {
break;
}
}
// getchar();
return 0;
}
基于哈夫曼树的数据压缩算法
这个还是比较有趣，可以自己试试
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
string s;
string ans[1000];
unordered_map<char,int> mmp;
vector<tuple<int,int,int,int>>tre;//重，fa，l,r
bool cmp(pair<char,int> a,pair<char,int> b)
{
return a.first>b.first;
}
void dfs(int x,string temp){
if(temp!="\0")
ans[x] =temp ;
int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
auto te1 = tre[x];
tie(w1,f1,l1,r1) = te1;
if(l1!=-1)dfs(l1,temp+'0');
if(r1!=-1)dfs(r1,temp+'1');
}
void solve(){
int m,n,idx = 0;
tre.clear();
memset(ans,'\0',sizeof(ans));
priority_queue<pi,vector<pi>,greater<pi>>mq;
mmp.clear();
for(auto j:s)mmp[j]++;
vector<pair<int,int> > ss;
for(auto j:mmp){
ss.push_back({j.first,j.second});
}
sort(ss.begin(),ss.end());
for(auto j:ss){
char ch = j.first;
printf("%c:%d ",ch,j.second);
mq.push({j.second,idx++});
//建树
tre.push_back({j.second,-1,-1,-1});
}
cout<<endl;
//建立合成节点的树
while(mq.size()>1){
auto t1 = mq.top();mq.pop();
auto t2 = mq.top();mq.pop();
int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
tie(w1,id1) = t1;tie(w2,id2) =t2;
mq.push({w1+w2,idx++});
tre.push_back({w1+w2,-1,id1,id2});
//以后要牢记tuple一般用于无修改的操作
//不是说内容一旦过长就用tuple；
auto te1 = tre[id1],te2 = tre[id2];
tie(w1,f1,l1,r1) = te1;
tie(w2,f2,l2,r2) = te2;
tre[id1] = {w1,idx-1,l1,r1};
tre[id2] = {w2,idx-1,l2,r2};
}
//输出树
jf0(idx){
int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
auto te1 = tre[j];
tie(w1,f1,l1,r1) = te1;
cout<<j+1<<" "<<w1<<" "<<f1+1<<" "<<l1+1<<" "<<r1+1<<endl;
}
//dfs编码树
dfs(idx-1,"\0");
if0(ss.size()){
char ch = ss[i].first;
cout<<ch<<":"<<ans[i]<<" ";
}
cout<<endl;
string res;
if0(ss.size()){
jf0(ss[i].second){
res+=ans[i];
}
}
cout<<res<<endl;
//解码，感觉会比较复杂
//奥！，dfs编码，那么我也应该是dfs解码，聪明like me
int now = idx-1;
for(auto j:res){
int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
auto te1 = tre[now];
tie(w1,f1,l1,r1) = te1;
if(l1==-1){//走到尽头辣,我们就可以输出当前节点的val，并且回溯到根节点
char ch = ss[now].first;
cout<<ch;now = idx-1;
auto te1 = tre[now];
tie(w1,f1,l1,r1) = te1;
}
if(j=='0'){
//向左边走、
now = l1;
}
else now = r1;
}
char ch = ss[now].first;
cout<<ch;now = idx-1;
cout<<endl;
}
int main(){
//ios::sync_with_stdio(false);
//cin.tie(nullptr);
//cout.tie(nullptr);
while (1)
{
cin>>s;
if(s=="0")break;
solve();
}
// getchar();
return 0;
}```
## 如何改变文本的样式
*强调文本* _强调文本_
**加粗文本** __加粗文本__
==标记文本==
~~删除文本~~
> 引用文本
H~2~O is是液体。
2^10^ 运算结果是 1024.
## 插入链接与图片
链接: [link](https://www.csdn.net/).
图片: 
带尺寸的图片: 
居中的图片: 
居中并且带尺寸的图片: 
当然，我们为了让用户更加便捷，我们增加了图片拖拽功能。
## 如何插入一段漂亮的代码片
去[博客设置](https://mp.csdn.net/console/configBlog)页面，选择一款你喜欢的代码片高亮样式，下面展示同样高亮的 `代码片`.
```javascript
// An highlighted block
var foo = 'bar';
基于二叉树的表达式求值算法
最傻逼的作业
#include<bits/stdc++.h>
using namespace std;
#define MAXSIZE 100
#define N 10
string S;
//运算符栈
unordered_map<char,int>mmp;
void init0(){
mmp['('] = mmp[')']=3;
mmp['*'] = mmp['/'] = 2;
mmp['+'] = mmp['-'] = 1;
}
struct node{
bool isnum;
int val;
char op;
node* L;
node* R;
};
stack<char> sta_op;
stack<node*> sta_num;
int verify(char ch)
{
//cout<<ch<<endl;
if(ch=='+')
return 0;
else if(ch=='-')
return 1;
else if(ch=='*')
return 2;
else if(ch=='/')
return 3;
else if(ch=='(')
return 4;
else if(ch==')')
return 5;
else
{
return 100;
}
}
void push_num(int& idx){
int j = idx;
int temp = 0;
while(S[j]>='0'&&S[j]<='9'){
temp*=10;
temp+=S[j]-'0';
j++;
}
node* te = new(node);
te->isnum=1;
te->val=temp;
sta_num.push(te);
idx = j-1;
}
int judge(int a,char e,int b)
{
//cout<<e<<endl;
if(e=='+')
return a+b;
else if(e=='-')
return a-b;
else if(e=='*')
return a*b;
else if(e=='/')
return a/b;
else
{
cout<<"error!"<<endl;
return -1;
}
}
void func1(int& idx){
char ch = S[idx];
if(ch=='('||ch=='+'||ch=='-')
sta_op.push(ch);//of course +-不需要考虑运算顺序
else if(ch == '*'||ch=='/'){
if(S[idx+1]>='0'&&S[idx+1]<='9'){
//*/在后面跟上的是数字的情况下，一定是可以直接进行运算的
//那么不妨在这种情况下面直接开算
int j = idx+1;
int temp = 0;
while(S[j]>='0'&&S[j]<='9'){
temp*=10;
temp+=S[j]-'0';
j++;
}
node* te = new(node);
te->isnum=1;
te->val=temp;
sta_num.push(te);
idx = j-1;
//
node* re = sta_num.top();
sta_num.pop();
node* la = sta_num.top();
sta_num.pop();
node* fa =new(node);
fa->isnum=0;
fa->op = ch;
fa->L=la;
fa->R = re;
sta_num.push(fa);
}
else{
sta_op.push(ch);//只能等后面在来运算辣
//就是等下一个左括号运算完成并出栈的时候来运算此符号
}
}else if(ch==')'){
while(sta_op.top()!='('){
node* re = sta_num.top();
sta_num.pop();
node* la = sta_num.top();
sta_num.pop();
node* fa =new(node);
fa->isnum=0;
fa->op = sta_op.top();
fa->L=la;
fa->R = re;
sta_num.push(fa);
sta_op.pop();
}
sta_op.pop();
if(sta_op.top()=='*'||sta_op.top()=='/'){//
//这个时候就要运算109行处的*/法辣
node* re = sta_num.top();
sta_num.pop();
node* la = sta_num.top();
sta_num.pop();
node* fa =new(node);
fa->isnum=0;
fa->op = sta_op.top();
fa->L=la;
fa->R = re;
sta_num.push(fa);
sta_op.pop();
}
}
}
int dfs(node* root){
if(root->isnum) return root->val;
int l = dfs(root->L),r=dfs(root->R);
return judge(l,root->op,r);
}
int func(){
cin>>S;
if(S=="=")return 0;
for(int i=0;S[i]!='\0';i++)
{
if(S[i]>='0'&&S[i]<='9')
push_num(i);
else
func1(i);
}
//*/法已经算完辣，+-法没有顺序可言，但是为了减少代码量就不优化辣。
while(!sta_op.empty()){
node* re = sta_num.top();
sta_num.pop();
node* la = sta_num.top();
sta_num.pop();
node* fa =new(node);
fa->isnum=0;
fa->op = sta_op.top();
fa->L=la;
fa->R = re;
sta_num.push(fa);
sta_op.pop();
}
node* root = sta_num.top();
cout<<dfs(root)<<endl;
sta_num.pop();
return 1;
}
int main()
{
int ti=1;
while(ti){
cout<<"\n请输入中缀表达式的算数表达式\n仅输入一个 = 时结束程序\n";
if(!func())break;
}
return 0;
}
基于 Dijsktra 算法的最短路
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
int e[200],h[200],ne[200],w[200],idx,idx1;
unordered_map<string ,int> mmp;
string xiufu[200];
int dist[200];
int pre[200];
int m,n,c;
void add(int a,int b,int c){
e[idx] = b;
w[idx] = c;
ne[idx] = h[a];
h[a] = idx++;
}
void dijsktra(int x){
memset(dist,inf,sizeof(dist));
priority_queue<pi> mq;
mq.push(make_pair(0,x));
dist[x]=0;
while(!mq.empty()){
int j = mq.top().second;
int distance = mq.top().first;
mq.pop();
for(int p = h[j];~p;p = ne[p]){
int q = e[p];//q是终点
if(distance+w[p]<dist[q]){
dist[q] = distance+w[p];
mq.push(make_pair(dist[q],q));
pre[q] = j;
}
}
}
}
void solve(){
string s;
string a,b;
idx1 = 1;
idx = 0;
mmp.clear();
if0(n)cin>>s;
memset(h,-1,sizeof(h));
if0(m){
cin>>a>>b>>c;
if(mmp[a]==0){
xiufu[idx1] = a;
mmp[a] = idx1++;
}
if(mmp[b]==0){
xiufu[idx1] = b;
mmp[b] = idx1++;
}
add(mmp[a],mmp[b],c);
}
cin>>a>>b;
memset(pre,-1,sizeof(pre));
dijsktra(mmp[a]);
stack<string>wedg;
int en = mmp[b];
while(en!=-1){
wedg.push(xiufu[en]);
en = pre[en];
}
cout<<dist[mmp[b]]<<endl;
while(!wedg.empty()){
cout<<wedg.top()<<" ";
wedg.pop();
}
cout<<endl;
}
int main(){
//ios::sync_with_stdio(false);
//cin.tie(nullptr);
//cout.tie(nullptr);
while(1){
cin>>n>>m;
if(m==0&&n==0)break;
solve();
}
return 0;
}
基于广度优先搜索的六度空间
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
vector<int> wedg[10005];
bool marked[10005];
int frend;
int m,n;
queue<pi>mq;
void bfs(){
while(!mq.empty()){
auto j = mq.front();
mq.pop();
if(marked[j.first]||j.second>6)continue;
marked[j.first]=1;
frend++;
for(auto t:wedg[j.first]){
if(!marked[t])mq.push(make_pair(t,j.second+1));
}
}
}
void solve(){
if0(m){
int a,b;
cin>>a>>b;
wedg[a].push_back(b);
wedg[b].push_back(a);
}
if1(n){
double ans;
memset(marked,0,sizeof(marked));
frend = 0;
mq.push({i,0});
bfs();
ans = 1.*(frend)/n;
printf("%d:%.2f%\n",i,ans*100);
}
if1(n)wedg[i].clear();
}
int main(){
while(1){
cin>>n>>m;
if(n==0&&m==0)return 0;
solve();
}
// getchar();
return 0;
}
排序算法的实现与分析
归并法
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
struct node{
string name;
double grade;
};
bool cmp(node a,node b){
return a.grade>b.grade;
}
void hebin(vector<node>& da,int l,int r,int mid){
node* temp = new node[r-l+2];
int i = l,j = mid+1,k=0;
while(i<=mid&&j<=r){
if(cmp(da[i],da[j])){
temp[k++] = da[i++];
}else{
temp[k++] = da[j++];
}
}
while(i<=mid)
temp[k++] = da[i++];
while(j<=r) temp[k++] = da[j++];
k=0;
for(int i = l; i<=r;)da[i++] = temp[k++];
delete []temp;
}
void guibin(int l,int r,vector<node>& da){
if(l<r){
int mid = l+r>>1;
guibin(l,mid,da);
guibin(mid+1,r,da);
hebin(da,l,r,mid);
}
}
int main(){
string s;
double grade;
vector<node> date;
while(1){
cin>>s>>grade;
if(s=="0"&&grade==0)break;
node te ;
te.grade = grade;
te.name = s;
date.push_back(te);
}
guibin(0,date.size()-1,date);
if0(date.size()){
cout<<date[i].name<<" ";
printf("%.2f\n",date[i].grade);
}
return 0;
}
![Kitex踩坑 [Error] KITEX: processing request error,i/o timeout](https://img-blog.csdnimg.cn/cb3a490b2ba943d5b4e6edb18a1b1095.png)
