STL算法刷题

2022-08-21

刷题遇到的一些小技巧

将输入大写转小写

string sentenct;
getline(cin, sentenct);
// up -> low
transform(sentenct.begin(),sentenct.end(),sentenct.begin(),::tolower);

string按空格分离字符串

string str;
getline(cin, str);
istringstream in(str);
while (in >> str) {
    keywordsToId[str].insert(id);
}

vector二维向量初始化
下面定义的是行为r，列为c的二维数组
1
vector<vector<int>> ans(r, vector<int>(c));

A1047 Student List for Course

思路：用vector 直接存储数据

#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
const int N = 400010;
const int K = 2510;

int n, k;
vector<string> Class[N];

bool cmp(string a, string b){
    return a < b;
}

int main (int argc, char *argv[])
{
    scanf("%d %d", &n, &k);
    string name;
    int c, cla;
    for (int i = 0; i < n; i++) {
        cin >> name >> c;
        for (int j = 0; j < c; j++) {
            scanf("%d", &cla);
            Class[cla].push_back(name);
        }
    }
    for (int i = 1; i <= k; i++) {
        printf("%d %lu\n", i, Class[i].size());
        sort(Class[i].begin(), Class[i].end(), cmp);
        for (int j = 0; j < Class[i].size(); j++) {
            // cout << Class[i][j] << endl;
            printf("%s\n", Class[i][j].c_str());
        }
    }
    return 0;
}

注意：要用printf作为输出，不然会超时。

A1063 Set Similarity

思路：set存储不重复值，交集并集运算。

#include <stdio.h>
#include <algorithm>
#include <set>
using namespace std;
const int N = 60;

set<int> sim[N];
double similaruty(int a, int b){
    set<int> add;
    add = sim[a];
    for (set<int>::iterator it = sim[b].begin(); it != sim[b].end(); it++) {
        add.insert(*it);
    }
    int all = sim[a].size() + sim[b].size();
    int Union = add.size();
    int intersection = all - Union;
    double ans = 1.0 * intersection / Union;
    return ans * 100;
}

int main (int argc, char *argv[])
{
    int n, m;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &m);
        int value;
        for (int j = 0; j < m; j++) {
            scanf("%d", &value);
            sim[i].insert(value);
        }
    }
    int k;
    int query1, query2;
    scanf("%d", &k);
    for (int i = 0; i < k; i++) {
        scanf("%d%d", &query1, &query2);
        double ans = similaruty(query1, query2);
        printf("%.1f%%\n", ans);
    }
    return 0;
}

一次过

1060 Are They Equal

思路：太难了，细节直接拉满，典型的string题，没什么套路

#include <cstdio>
#include <iterator>
#include <stdio.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 110;

int n;
// true 是小数 
bool cutZero(string& str){
    int len = str.size();
    for (int i = 0; i < len; i++) {
        if (str[0] == '0') {
            str.erase(str.begin());
        }
        else if(str[0] == '.'){
            str.erase(str.begin());
            return true;
        }
        else {
            return false;
        }
    }
    //0000.00
    return false;
}

void getString(string& str, int& e){
    if (cutZero(str)) {
        int len = str.size();
        for (int i = 0; i < len; i++) {
            if (str[0] == '0') {
                e--;
                str.erase(str.begin());
            }
        }
        // 0000.00
        if (str.size() == 0) {
            str = "0";
            e = 0;
        }
        while (str.size() <= n) {
            str += '0';
        }
        str = str.substr(0, n);
    }
    else {
        int len = str.size();
        for (int i = 0; i < len; i++) {
            if (str[i] != '.') {
                e++;
            }
            else if (str[i] == '.') {
                str.erase(i, 1);
                break;
            }
        }
        if (str.size() == 0) {
            str = "0";
            e = 0;
        }
        //补齐位数
        while (str.size() <= n) {
            str += '0';
        }
        str = str.substr(0, n);
    }
}

int main (int argc, char *argv[])
{
    string str1, str2;
    cin >> n >> str1 >> str2;
    int e1 = 0, e2 = 0;
    getString(str1, e1);
    getString(str2, e2);
    if (str1 == str2 && e1 == e2) {
        cout << "YES 0." << str1 << "*10^" << e1 << endl;
    }
    else {
        cout << "NO 0." << str1 << "*10^" << e1;
        cout << " 0." << str2 << "*10^" << e2;
        cout << endl;
    }
    return 0;
}

这里犯了一个极大的错误，str.size() , 但我的str是引用，导致我的每次earse都会减小size，循环会提前退出。以及全是0的判断，位数补齐要补零，不然也会出错。

1100 Mars Numbers

思路：直接打表，string和int的对应关系。

#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
const int N = 170;

string tenDigit[13] = {"tret","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
string unitDigit[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
//打表

map<string, int> stringToInt;
map<int, string> intToString;
void init(){
    string ans;
    for (int i = 0; i < 13; i++) {
        for (int j = 0; j < 13; j++) {
            if (i == 0 && j == 0) {
                ans = "tret";
                stringToInt[ans] = 13 * i + j;
                intToString[13*i+j] = ans;
            }
            else if (i == 0) {
                ans = unitDigit[j];
                stringToInt[ans] = 13 * i + j;
                intToString[13*i+j] = ans;
            }
            else if (j == 0) {
                ans = tenDigit[i];
                stringToInt[ans] = 13 * i + j;
                intToString[13*i+j] = ans;
            }
            else{
                ans = tenDigit[i] + " " + unitDigit[j];
                stringToInt[ans] = 13 * i + j;
                intToString[13*i+j] = ans;
            }
        }
    }
}
int main (int argc, char *argv[])
{
    init();
    int n;
    scanf("%d", &n);
    getchar();
    string str;
    for (int i = 0; i < n; i++) {
        int num = 0;
        getline(cin, str);
        if (str[0] >= '0' && str[0] <= '9') {
            for (int i = 0; i < str.length(); i++) {
                num = num * 10 + str[i] - '0'; 
            }
            cout << intToString[num] << endl;
        }
        else {
            cout << stringToInt[str] << endl;
        }
    }
    return 0;
}

注意，13 26 这样的数只会出现高位，低位的零会舍弃。

A1054 The Dominant Color

思路：map<int, int> 映射像素和数量的关系，当超过半数，则输出。```

#include <stdio.h>
#include <algorithm>
#include <map>
using namespace std;

int main (int argc, char *argv[])
{
    int m, n;
    scanf("%d%d", &m, &n);
    int half = n * m / 2;
    int value;
    map<int, int> ans;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%d", &value);
            ans[value]++;
            if (ans[value] > half) {
                printf("%d\n", value);
                return 0;
            }
        }
    }
    return 0;
}

思路好难想到。

A1071 Speech Patterns

思路：判断有效单词，并截取，map进行统计。

#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
#include <map>
using namespace std;

int main (int argc, char *argv[])
{
    string sentenct;
    getline(cin, sentenct);
    // up -> low
    transform(sentenct.begin(),sentenct.end(),sentenct.begin(),::tolower);
    // sentenct += "  ";
    int start = 0, len = 0;
    map<string, int> ans;
    string str;
    for (int i = 0; i < sentenct.size(); i++) {
        if ((sentenct[i] >= '0' && sentenct[i] <= '9') || (sentenct[i] >= 'a' && sentenct[i] <= 'z')) {
            len++;
        }
        // method2 up -> low
        /* else if (sentenct[i] >= 'A' && sentenct[i] <= 'Z') {
            sentenct[i] = tolower(sentenct[i]);
            len++;
        } */
        else {
            if (len != 0) {
                str = sentenct.substr(start, len);
                if (ans.find(str) != ans.end()) {
                    ans[str]++;
                }
                else {
                    ans[str] = 1;
                }
            }
            start = i + 1;
            len = 0;
        }
    }
    int max = 0;
    string res;
    map<string, int>::iterator it;
    for (it = ans.begin(); it != ans.end(); it++) {
        if (it->second > max) {
            max = it->second;
            res = it->first;
        }
    }
    cout << res << " " << max << endl;
    return 0;
}

注意：开始时加了sentence += “”是为了处理下面for循环的边界问题，不然测试点4过不了。

A1022 Digital Library

思路:建立map映射，查询就变成了o(1).

#include <cstdio>
#include <sstream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <map>
#include <iostream>
#include <set>
using namespace std;
const int N = 1e4 + 10;

map<string, set<int> > titleToId;
map<string, set<int> > authorToId;
map<string, set<int> > keywordsToId;
map<string, set<int> > publisherToId;
map<string, set<int> > yearToId;

int main (int argc, char *argv[])
{
    int n;
    scanf("%d", &n);
    int id;
    for (int i = 0; i < n; i++) {
        scanf("%d\n", &id);
        string str;
        for (int j = 0; j < 5; j++) {
            getline(cin, str);
            if (j == 0) {
                titleToId[str].insert(id);
            }
            else if (j == 1) {
                authorToId[str].insert(id);
            }
            else if (j == 2) {
                istringstream in(str);
                while (in >> str) {
                    keywordsToId[str].insert(id);
                }
            }
            else if (j == 3) {
                publisherToId[str].insert(id);
            }
            else {
                yearToId[str].insert(id);
            }
        }
    }
    int m;
    scanf("%d", &m);
    int indexQuery = 0;
    string contentQuery;
    for (int i = 0; i < m; i++) {
        scanf("%d: ", &indexQuery);
        getline(cin, contentQuery);
        cout << indexQuery << ": " << contentQuery << endl;
        if (indexQuery ==  1) {
            if (titleToId.find(contentQuery) == titleToId.end()) {
                printf("Not Found\n");
            }
            else {
                for (set<int>::iterator it = titleToId[contentQuery].begin(); it != titleToId[contentQuery].end(); it++) {
                    printf("%07d\n", *it);
                }
            }
        }
        else if (indexQuery ==  2) {
            if (authorToId.find(contentQuery) == authorToId.end()) {
                printf("Not Found\n");
            }
            else {
                for (set<int>::iterator it = authorToId[contentQuery].begin(); it != authorToId[contentQuery].end(); it++) {
                    printf("%07d\n", *it);
                }
            }
        }
        else if (indexQuery ==  3) {
            if (keywordsToId.find(contentQuery) == keywordsToId.end()) {
                printf("Not Found\n");
            }
            else {
                for (set<int>::iterator it = keywordsToId[contentQuery].begin(); it != keywordsToId[contentQuery].end(); it++) {
                    printf("%07d\n", *it);
                }
            }
        }
        else if (indexQuery ==  4) {
            if (publisherToId.find(contentQuery) == publisherToId.end()) {
                printf("Not Found\n");
            }
            else {
                for (set<int>::iterator it = publisherToId[contentQuery].begin(); it != publisherToId[contentQuery].end(); it++) {
                    printf("%07d\n", *it);
                }
            }
        }
        else {
            if (yearToId.find(contentQuery) == yearToId.end()) {
                printf("Not Found\n");
            }
            else {
                for (set<int>::iterator it = yearToId[contentQuery].begin(); it != yearToId[contentQuery].end(); it++) {
                    printf("%07d\n", *it);
                }
            }
        }
    }
    return 0;
}

注意，用set可以直接得到排序的结果，用vector还要用sort。用int接受id的话，输出要前面补零，否则测试点3 4过不了。