STL专题

一、algorithm

1、sort

问题1：给你n个整数，请按从大到小的顺序输出其中前m大的数。

Input：每组测试数据有两行，第一行有两个数n,m(0<n,m<1000000)，第二行包含n个各不相同，且都处于区间[-500000,500000]的整数。
Output：对每组测试数据按从大到小的顺序输出前m大的数。
Sample Input：

5 3
3 -35 92 213 -644

Sample Output：

213 92 3

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

bool cmp(int a,int b){
    return a>b;
}

int main(){
    int num1,num2;
    while(cin>>num1>>num2){
        int a[];
        for(int i=;i<num1;i++){
            scanf("%d",&a[i]);
        }
        sort(a,a+num1,cmp);
        for(int i=;i<num2;i++){
            cout<<a[i]<<" ";
        }
        cout<<endl;
    }
    system("pause");
    return ;
}

问题2：

Welcome to HDU to take part in the first CCPC girls' competition!

As a pretty special competition, many volunteers are preparing for it with high enthusiasm.
One thing they need to do is blowing the balloons.

Before sitting down and starting the competition, you have just passed by the room where the boys are blowing the balloons. And you have found that the number of balloons of different colors are strictly different.

After thinking about the volunteer boys' sincere facial expressions, you noticed that, the problem with more balloon numbers are sure to be easier to solve.

Now, you have recalled how many balloons are there of each color.
Please output the solving order you need to choose in order to finish the problems from easy to hard.
You should print the colors to represent the problems.

Input：The first line is an integer T which indicates the case number.
And as for each case, the first line is an integer n, which is the number of problems.
Then there are n lines followed, with a string and an integer in each line, in the i-th line, the string means the color of ballon for the i-th problem, and the integer means the ballon numbers.

It is guaranteed that:
T is about 100.
1≤n≤10.
1≤ string length ≤10.
1≤ bolloon numbers ≤83.(there are 83 teams :p)
For any two problems, their corresponding colors are different.
For any two kinds of balloons, their numbers are different.
Output：For each case, you need to output a single line.
There should be n strings in the line representing the solving order you choose.
Please make sure that there is only a blank between every two strings, and there is no extra blank.
Sample Input：

3
3
red 1
green 2
yellow 3
1
blue 83
2
red 2
white 1

Sample Output：

yellow green red
blue
red white

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
using namespace std;

struct ss{
    string col;
    int val;
}num[];

bool cmp(ss a,ss b){
    //return a.val>b.val;
    if(a.val == b.val){
        return a.col>b.col;
    }else{
        return a.val>b.val;
    }
}

int main(){
    int t;
    cin>>t;
    getchar();
    while(t--){
        int n;
        cin>>n;
        for(int i=;i<n;i++){
            cin>>num[i].col;
            cin>>num[i].val;
        }
        sort(num,num+n,cmp);
        for(int i=;i<n-;i++){
            cout<<num[i].col<<" ";
        }
        cout<<num[n-].col<<endl;
    }
    system("pause");
    return ;
}

问题3：插入排序

代码：

#include <cstdio>
#include <iostream>
using namespace std;

int main(){
    int temp[];
    for(int i=;i<=;i++){
        cin>>temp[i];
    }
    for(int i=;i<=;i++){
        int t = temp[i];
        int j = i;
        while(j>&&t>temp[j-]){
            temp[j] = temp[j-];
            j--;
        }
        temp[j] = t;
    }
    for(int i=;i<=;i++){
        cout<<temp[i]<<" ";
    }
    cout<<endl;
    return ;
}

2、二分（有序序列）

eg:{1,2,2,3,3,3,5,5,5,5}

lower_bound：第一个满足大于等于x的元素的位置      eg:3

upper_bound：第一个满足大于x的元素的位置             eg:6

问题：给一串n个单调递增的数，有q次询问>=x且<=y的数有多少个。

Sample Input：

1
5 3
1 4 6 8 10
0 5
6 10
7 100000

Sample Output：

Case 1:

2

3

3

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
using namespace std;

int main(){
    int t;
    scanf("%d",&t);
    for(int j=;j<=t;j++){
        printf("Case %d:\n",j);
        int n,m;
        scanf("%d%d",&n,&m);
        int temp[];
        for(int i=;i<n;i++){
            scanf("%d",&temp[i]);
        }
        while(m--){

            int q1,q2;
            scanf("%d%d",&q1,&q2);
            int begin = (int)(lower_bound(temp,temp+n,q1)-temp);
            int end   = (int)(upper_bound(temp,temp+n,q2)-temp);
            printf("%d\n",end-begin);
        }
    }
    return ;
}

3、归并排序与逆序对个数

 

4、next_permutation

问题：全排列

Sample Input

6 4
11 8

Sample Output

1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

int main(){
    int n,m;
    while(cin>>n>>m){
        int a[];
        for(int i=;i<n;i++){
            a[i] = i+;
        }
        int p=;
        do{
            p++;
            if(p==m){
                for(int i=;i<n-;i++){
                    cout<<a[i]<<" ";
                }
                cout<<a[n-]<<endl;
                break;
            }
        }while(next_permutation(a,a+n));
    }
    system("pause");
    return ;
}

二、stack

问题1：

Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.

Input：The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single line with several words. There will be at most 1000 characters in a line.
Output：For each test case, you should output the text which is processed.
Sample Input：

3
olleh !dlrow
m'I morf .udh
I ekil .mca

Sample Output：

hello world!
I'm from hdu.
I like acm.

代码：

#include<iostream>
#include<algorithm>
#include<stack>
#include<string>
#include<cstdio>
using namespace std;

int main(){
    string s;
    int t;
    cin>>t;
    getchar();
    while(t--){
        getline(cin,s);
        int len = (int)s.size();
        stack<char>st;
        for(int i =;i<len;i++){
            if(s[i] != ' '){
                st.push(s[i]);
            }
            if(s[i] == ' '||i == len-){
                while(!st.empty()){
                    printf("%c",st.top());//获得栈顶元素
                    st.pop();//弹出栈顶元素
                }
                if(s[i] == ' '){
                    cout<<" ";
                }
            }
        }
        cout<<endl;
    }
    system("pause");
    return ;
}

问题2：

读入一个只包含 +, -, *, / ,（,）的非负整数计算表达式，计算该表达式的值。

Input：测试输入包含若干测试用例，每个测试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中只有0时输入结束，相应的结果不要输出。
Output：对每个测试用例输出1行，即该表达式的值，精确到小数点后2位。
Sample Input：

1 + 2
4 + 2 * 5 - 7 / 11
0

Sample Output：

3.00
13.36

代码1（复杂，适用于所有）：

#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<cstring>
#include<cstdio>
#include<cstdlib>
using namespace std;
//cout<<"AAAA"<<endl;

struct node{
    double num;
    char op;
    bool flag;
};

stack<node> s;
queue<node> q;
map<char,int> op;

void change(string str);
double cal();

string str;
int main(){
    op['+'] = op['-'] = ;
    op['*'] = op['/'] = ;
    while(getline(cin,str),str!=""){
        string::iterator it;
        for(it=str.end();it!=str.begin();it--){
            if(*it == ' '){
                str.erase(it);
            }
        }
        while(!s.empty()){
            s.pop();
        }
        change(str);
        printf("%0.2f\n",cal());
    }
    system("pause");
    return ;
}

void change(string str){//前缀变后缀
    node temp;
    for(int i=;i<(int)str.size();){
        if(str[i]=='('){
            temp.flag = false;
            temp.op = str[i];
            s.push(temp);
            i++;
        }else if(str[i]==')'){
            while(!s.empty()&&s.top().op!='('){
                q.push(s.top());
                s.pop();
            }
            s.pop();
            i++;
        }else if(str[i]>=''&&str[i]<=''){
            temp.flag = true;
            temp.num = str[i] - '';
            i++;
            while(i<(int)str.size()&&str[i]>=''&&str[i]<=''){
                temp.num = temp.num*+(str[i]-'');
                i++;
            }
            q.push(temp);
        }else{
            temp.flag = false;
            while(!s.empty()&&op[str[i]]<=op[s.top().op]){
                q.push(s.top());
                s.pop();
            }
            temp.op = str[i];
            s.push(temp);
            i++;
        }
    }
    while(!s.empty()){
        q.push(s.top());
        s.pop();
    }
}

double cal(){//计算后缀表达式:1 1 +
    double temp1,temp2;
    node cur,temp;
    while(!q.empty()){
        cur = q.front();
        q.pop();
        if(cur.flag==true){
            s.push(cur);
        }else{
            temp2 = s.top().num;
            s.pop();
            temp1 = s.top().num;
            s.pop();
            temp.flag = true;
            if(cur.op == '+'){
                temp.num = temp1+temp2;
            }else if(cur.op == '-'){
                temp.num = temp1-temp2;
            }else if(cur.op == '*'){
                temp.num = temp1*temp2;
            }else{
                temp.num = temp1/temp2;
            }
            s.push(temp);
        }
    }
    return s.top().num;
}

代码2（适用于带空格但没有括号的表达式）：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
using namespace std;

int main(){
    double num;
    while(cin>>num){
        stack<double> st;
        char ch = getchar();
        if(ch!=' '&&num==){
            break;
        }
        st.push(num);
        double n;
        char c,s;
        while(scanf("%c %lf%c",&c,&n,&s)!=EOF){
            if(c=='+'){
                st.push(n);
            }else if(c=='-'){
                st.push(-*n);
            }else if(c=='*'){
                st.top() = st.top()*n;
            }else if(c=='/'){
                st.top() = st.top()/n;
            }
            if(s!=' '){
                break;
            }
        }
        double sum = 0.0;
        while(!st.empty()){
            sum+=st.top();
            st.pop();
        }
        printf("%.2f\n",sum);
    }
    return ;
}

代码3（适用于不带空格和括号的表达式）：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
using namespace std;

int main(){
    double num;
    while(cin>>num){
        stack<double> st;
        if(num==){
            break;
        }
        st.push(num);
        double n;
        char c;
        while(scanf("%c",&c)!=EOF){
            if(c=='\n'){
                break;
            }
            cin>>n;
            if(c=='+'){
                st.push(n);
            }else if(c=='-'){
                st.push(-*n);
            }else if(c=='*'){
                st.top() = st.top()*n;
            }else if(c=='/'){
                st.top() = st.top()/n;
            }
        }
        double sum = 0.0;
        while(!st.empty()){
            sum+=st.top();
            st.pop();
        }
        printf("%.2f\n",sum);
    }
    return ;
}

代码3（适用于不带空格和括号的表达式且用数组模拟栈）：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
using namespace std;

int main(){
    int num;
    int temp[],p=;
    int sum;
    while(cin>>num){
        sum = ;
        memset(temp,,sizeof(temp));
        temp[p++] = num;
        int n;
        char c;
        while(scanf("%c",&c)!=EOF){
            if(c=='\n'){
                break;
            }
            cin>>n;
            if(c=='+'){
                temp[p] = n;
                p++;
            }else if(c=='-'){
                temp[p] = -*n;
                p++;
            }else if(c=='*'){
                p--;
                temp[p]= temp[p]*n;
                p++;
            }else if(c=='/'){
                p--;
                temp[p]= temp[p]/n;
                p++;
            }
        }
        for(int i=;i<p;i++){
            sum+=temp[i];
        }
        printf("%d\n",sum);
    }
    return ;
}

三、queue

问题：

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

Input:The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

Output:For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

Sample Input:

4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT

Sample Output:

1
2
2
1
1
2
None
2
3

代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<string>
#include<cstdio>
using namespace std;

int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        string s1,s2;
        cin>>s1;
        if(s1 == "FIFO"){
            queue<int> qu;
            while(n--){
                cin>>s2;
                if(s2 == "IN"){
                    int num;
                    cin>>num;
                    qu.push(num);
                }else{
                    if(qu.empty()){
                        cout<<"None"<<endl;
                    }else{
                        cout<<qu.front()<<endl;
                        qu.pop();
                    }
                }
            }
        }else{
            stack<int> st;
            while(n--){
                cin>>s2;
                if(s2 == "IN"){
                    int num;
                    cin>>num;
                    st.push(num);
                }else{
                    if(st.empty()){
                        cout<<"None"<<endl;
                    }else{
                        cout<<st.top()<<endl;
                        st.pop();
                    }
                }
            }
        }
    }
    system("pause");
    return ;
}

四、set（内部自动有序、不含重复）

问题：

有一群人，打乒乓球比赛，两两捉对撕杀，每两个人之间最多打一场比赛。
球赛的规则如下：
如果A打败了B，B又打败了C，而A与C之间没有进行过比赛，那么就认定，A一定能打败C。
如果A打败了B，B又打败了C，而且，C又打败了A，那么A、B、C三者都不可能成为冠军。
根据这个规则，无需循环较量，或许就能确定冠军。你的任务就是面对一群比赛选手，在经过了若干场撕杀之后，确定是否已经实际上产生了冠军。

Input输入含有一些选手群，每群选手都以一个整数n(n<1000)开头，后跟n对选手的比赛结果，比赛结果以一对选手名字（中间隔一空格）表示，前者战胜后者。如果n为0，则表示输入结束。
Output对于每个选手群，若你判断出产生了冠军，则在一行中输出“Yes”，否则在一行中输出“No”。
Sample Input

3
Alice Bob
Smith John
Alice Smith
5
a c
c d
d e
b e
a d
0

Sample Output

Yes
No

代码：

#include<iostream>
#include<algorithm>
#include<set>
#include<string>
#include<cstdio>
using namespace std;

int main(){
    int t;
    while(cin>>t){
        if(t == ){
            break;
        }
        set<string> st1,st2;
        string s1,s2;
        while(t--){
            cin>>s1>>s2;
            st1.insert(s1);
            st1.insert(s2);
            st2.insert(s2);
        }
        if(st1.size()-st2.size() != ){
            cout<<"No"<<endl;
        }else{
            cout<<"Yes"<<endl;
        }
    }
    system("pause");
    return ;
}

五、迭代器

问题：

参加过上个月月赛的同学一定还记得其中的一个最简单的题目，就是{A}+{B}，那个题目求的是两个集合的并集，今天我们这个A-B求的是两个集合的差，就是做集合的减法运算。（当然，大家都知道集合的定义，就是同一个集合中不会有两个相同的元素，这里还是提醒大家一下）

呵呵，很简单吧？

Input每组输入数据占1行,每行数据的开始是2个整数n(0<=n<=100)和m(0<=m<=100),分别表示集合A和集合B的元素个数，然后紧跟着n+m个元素，前面n个元素属于集合A，其余的属于集合B. 每个元素为不超出int范围的整数,元素之间有一个空格隔开.
如果n=0并且m=0表示输入的结束，不做处理。

Output针对每组数据输出一行数据,表示A-B的结果,如果结果为空集合，则输出“NULL”,否则从小到大输出结果,为了简化问题，每个元素后面跟一个空格.
Sample Input

3 3 1 2 3 1 4 7
3 7 2 5 8 2 3 4 5 6 7 8
0 0

Sample Output

2 3
NULL

代码：

#include<iostream>
#include<algorithm>
#include<set>
#include<string>
#include<cstdio>
using namespace std;

int main(){
    int num1,num2;
    while(cin>>num1>>num2){
        if(num1 == num2&&num1 == ){
            break;
        }
        int num;
        set<int> st;
        for(int i=;i<num1;i++){
            cin>>num;
            st.insert(num);
        }
        for(int i=;i<num2;i++){
            cin>>num;
            if(st.find(num) != st.end()){
                st.erase(num);
            }
        }
        if(st.size() == ){
            cout<<"NULL"<<endl;
            continue;
        }
        set<int>::iterator it;
        for(it=st.begin();it!=st.end();it++){
            cout<<*it<<" ";
        }
        cout<<endl;
    }
    system("pause");
    return ;
}

六、map(键值对)

 问题：

Every girl likes shopping,so does dandelion.Now she finds the shop is increasing the price every day because the Spring Festival is coming .She is fond of a shop which is called "memory". Now she wants to know the rank of this shop's price after the change of everyday.

InputOne line contians a number n ( n<=10000),stands for the number of shops.
Then n lines ,each line contains a string (the length is short than 31 and only contains lowercase letters and capital letters.)stands for the name of the shop.
Then a line contians a number m (1<=m<=50),stands for the days .
Then m parts , every parts contians n lines , each line contians a number s and a string p ,stands for this day ,the shop p 's price has increased s.
OutputContains m lines ,In the ith line print a number of the shop "memory" 's rank after the ith day. We define the rank as :If there are t shops' price is higher than the "memory" , than its rank is t+1.Sample Input

3
memory
kfc
wind
2
49 memory
49 kfc
48 wind
80 kfc
85 wind
83 memory

Sample Output

1
2

代码：

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstdio>
using namespace std;

int main(){
    int t;
    while(cin>>t){
        for(int i=;i<t;i++){
            string s;
            cin>>s;
        }
        int n;
        cin>>n;
        map<string,int> shop;
        while(n--){
            for(int i=;i<t;i++){
                string name;
                int price;
                cin>>price;
                cin>>name;
                shop[name]+=price;
            }
            map<string,int>::iterator it;
            int mas = ;
            for(it=shop.begin();it!=shop.end();it++){
                if(it->second>shop["memory"]){
                    mas++;
                }
            }
            cout<<mas<<endl;
        }
    }
    system("pause");
    return ;
}

七、priority_queue(优先队列)

 问题：

看病要排队这个是地球人都知道的常识。
不过经过细心的0068的观察，他发现了医院里排队还是有讲究的。0068所去的医院有三个医生（汗，这么少）同时看病。而看病的人病情有轻重，所以不能根据简单的先来先服务的原则。所以医院对每种病情规定了10种不同的优先级。级别为10的优先权最高，级别为1的优先权最低。医生在看病时，则会在他的队伍里面选择一个优先权最高的人进行诊治。如果遇到两个优先权一样的病人的话，则选择最早来排队的病人。

现在就请你帮助医院模拟这个看病过程。

Input输入数据包含多组测试，请处理到文件结束。
每组数据第一行有一个正整数N(0<N<2000)表示发生事件的数目。
接下来有N行分别表示发生的事件。
一共有两种事件：
1:"IN A B",表示有一个拥有优先级B的病人要求医生A诊治。(0<A<=3,0<B<=10)
2:"OUT A",表示医生A进行了一次诊治，诊治完毕后，病人出院。(0<A<=3)

Output对于每个"OUT A"事件，请在一行里面输出被诊治人的编号ID。如果该事件时无病人需要诊治，则输出"EMPTY"。
诊治人的编号ID的定义为：在一组测试中，"IN A B"事件发生第K次时，进来的病人ID即为K。从1开始编号。
Sample Input

7
IN 1 1
IN 1 2
OUT 1
OUT 2
IN 2 1
OUT 2
OUT 1
2
IN 1 1
OUT 1

Sample Output

2
EMPTY
3
1
1

代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<cstdio>
using namespace std;

struct node{
    int value,id;
    bool operator<(const node &b)const{
        if(value == b.value){
            return id>b.id;
        }else{
            return value<b.value;
        }
    }
};

int main(){
    int t;
    while(cin>>t){
        node pep;
        priority_queue<node> q[];
        int time = ;
        string s;
        while(t--){
            cin>>s;
            if(s == "IN"){
                int a;
                cin>>a>>pep.value;
                pep.id = time;
                time++;
                q[a-].push(pep);
            }else{
                int a;
                cin>>a;
                if(q[a-].empty()){
                    cout<<"EMPTY"<<endl;
                }else{
                    cout<<q[a-].top().id<<endl;
                    q[a-].pop();
                }
            }
        }
    }
    system("pause");
    return ;
}

八、list

问题：

某部队进行新兵队列训练，将新兵从一开始按顺序依次编号，并排成一行横队，训练的规则如下：从头开始一至二报数，凡报到二的出列，剩下的向小序号方向靠拢，再从头开始进行一至三报数，凡报到三的出列，剩下的向小序号方向靠拢，继续从头开始进行一至二报数。。。，以后从头开始轮流进行一至二报数、一至三报数直到剩下的人数不超过三人为止。

Input本题有多个测试数据组，第一行为组数N，接着为N行新兵人数，新兵人数不超过5000。
Output共有N行，分别对应输入的新兵人数，每行输出剩下的新兵最初的编号，编号之间有一个空格。
Sample Input

2
20
40

Sample Output

1 7 19
1 19 37

代码1：

#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<cstdio>
#include<stack>
#include<list>
#include<cmath>
#include<cstring>//
#include<cstdlib>//
using namespace std;

int main(){
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        int k = ;
        list<int> blist;
        list<int>::iterator it;
        for(int i=;i<=n;i++){
            blist.push_back(i);
        }
        while(blist.size()>){
            int num =;
            for(it=blist.begin();it!=blist.end();){
                if(num++%k == ){
                    it = blist.erase(it);
                }else{
                    it++;
                }
            }
            if(k == ){
                k = ;
                continue;
            }else{
                k = ;
                continue;
            }
            //k==2?k=3:k=2;
        }
        for(it=blist.begin();it!=blist.end();it++){
            if(it!=blist.begin()){
                cout<<" ";
            }
            cout<<*it;
        }
        cout<<endl;
    }
    system("pause");
    return ;
}

代码2：