Daliy Algorithm (dp，模拟)-- day 80

Nothing to fear

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种一棵树最好的时间是十年前，其次是现在！

那些你早出晚归付出的刻苦努力，你不想训练，当你觉的太累了但还是要咬牙坚持的时候，那就是在追逐梦想，不要在意终点有什么，要享受路途的过程，或许你不能成就梦想，但一定会有更伟大的事情随之而来。 mamba out~

2020.5.16

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人一我十，人十我百，追逐青春的梦想，怀着自信的心，永不言弃！

PAT-A1022

经验:

1. 如果getline(cin,xxx)之前出现别的scanf / cin 需要一个
   getchar吸收掉之前的回车！因为之前的回车还保留在缓冲区内

2. PAT 将场上卡坑的数据比如输出几位数字7位必须输出前导0

3. 考虑每一个数据的对应特殊情况

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <string>

using namespace std;
#define SN std::string::npos

int n , m;
struct book{
    int id;
    string book_title;
    string author;
    string keywords;
    string publisher;
    string published_year;
};
vector<book> a;
bool cmp(book a,book b)
{
    return a.id < b.id;
}   
void query(int way,string key)
{
    bool flag = 0;
    printf("%d: ",way);cout << key << endl;
    for(int i = 0;i < a.size() ;i ++)
    {
        if(way == 1)
        {
            if(a[i].book_title == key)
            {
                printf("%07d
", a[i].id);
                flag = 1;
            }
        }
        if(way == 2)
        {
            if(a[i].author == key)
            {
                flag = 1;
                printf("%07d
", a[i].id);
            }
        }
        if(way == 3)
        {
            if(a[i].keywords.find(key) != SN)
            {
                flag = 1;
                printf("%07d
", a[i].id);
            }
        }
        if(way == 4)
        {
            if(a[i].publisher == key)
            {
                flag = 1;
                printf("%07d
", a[i].id);
            }
        }
        if(way == 5)
        {
            if(a[i].published_year == key)
            {
                flag = 1;
                printf("%07d
", a[i].id);
            }
        }
    }
    if(!flag)cout << "Not Found" << endl;
}
void input()
{
    cin >> n;
    int aa ;string b ,c ,d ,e,f;
    for(int i = 0;i < n ;i ++)
    {
        cin >> aa;getchar();getline(cin,b);
        getline(cin,c);getline(cin,d);
        getline(cin,e);cin >> f;
        a.push_back({aa,b,c,d,e,f});
    }
    sort(a.begin(),a.end(),cmp);
    cin >> m;
    int choice ;string key;
    for(int i = 0;i < m;i ++)
    {
        scanf("%d: ",&choice);
        getline(cin,key);
        query(choice , key);
    }
}
int main()
{
    input();
    return 0;
}

下面是dp复习

Constanze's Machine

通常遇到和前两种状态或者两种字符状态相关的最值问题，首先是动态规划问题，其次考虑斐波那契数列的变形。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cassert>
#include <string>
#include <iomanip>
#include <cmath>
#include <ctime>
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define lowbit(x) (x & -x)
using namespace std;
typedef long long ll;
const int MAX = 0x7ffffff;
const int mod = 1e9 + 7;
int t;

void slove()
{
    string s;
    cin >> s;
    s = '0' + s;
    int n = s.length();
    vector<ll> f(n , 0);
    f[0] = 1 , f[1] = 1;
    for(int i = 1; i <= n;i ++)
    {
        if(s[i] == 'm' || s[i] == 'w'){
            cout << 0 << endl;return;
        }
        f[i] = f[i-1];
        if(s[i] == 'u')
        {
            if(s[i-1] == 'u')f[i] = f[i-1] + f[i-2];
            else f[i] = f[i-1];
        }
        else if(s[i] == 'n')
        {
            if(s[i-1] == 'n')f[i] = f[i-1] + f[i-2];
            else f[i] = f[i-1];
        }
        f[i] = f[i] % mod;
    }
    cout << f[n] << endl;
}
int main()
{
#ifdef LOCAL
    auto start_time = clock();
    cerr << setprecision(3) << fixed; // 在iomanip中
#endif
    SIS;
    slove();
#ifdef LOCAL
    auto end_time = clock();
    cerr << "Execution time: " << (end_time - start_time) * (int)1e3 / CLOCKS_PER_SEC << " ms
";
#endif
}

Basketball Exercise

关键在于对于f3的状态的计算是从排除掉当前f1和f2所包含的阶段状态！

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cassert>
#include <string>
#include <iomanip>
#include <cmath>
#include <ctime>
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define lowbit(x) (x & -x)
using namespace std;
typedef long long ll;
const int MAX = 0x7ffffff;
int t;

void slove()
{
    int n;
    cin >> n;
    vector<ll> a(n + 1),b(n + 1);
    for(int i = 1;i <= n ;i ++)cin >> a[i];
    for(int i = 1;i <= n ;i ++)cin >> b[i];
    vector<ll> f1(n + 1),f2(n+1),f3(n+1);
    
    f1[1] = a[1];f2[1] = b[1];
    for(int i = 2;i <= n ;i ++)
    {
        f1[i] = max(max(f2[i-1],f3[i-1]) + a[i], a[i]);
        f2[i] = max(max(f1[i-1],f3[i-1]) + b[i], b[i]);
        f3[i] = max(f1[i-1],f2[i-1]);
    }
    cout << max(f1[n],f2[n]) << endl;
}
int main()
{
#ifdef LOCAL
    auto start_time = clock();
    cerr << setprecision(3) << fixed; // 在iomanip中
#endif
    SIS;
    slove();
#ifdef LOCAL
    auto end_time = clock();
    cerr << "Execution time: " << (end_time - start_time) * (int)1e3 / CLOCKS_PER_SEC << " ms
";
#endif
}

LG-P1435 回文字串

关键在于将问题转化为LCS问题去解决。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 1005;
char a[N] , b[N];
int f[N][N];
int main()
{
    scanf("%s",a + 1);
    int n = strlen(a + 1);

    for(int i = 1 ;i <= n; i++)b[i] = a[n-i+1];
    
    for(int i = 1;i <= n ;i ++)
    {
        for(int j = 1;j <= n ;j ++)
        {
            f[i][j] = 1;
            if(a[i] == b[j])f[i][j] = f[i-1][j-1] + 1;
            else f[i][j] = max(f[i-1][j],f[i][j-1]);
        }
    }
    cout << n - f[n][n] << endl;
    return 0;
}