线性dp

1.数字三角形。acwing 898.

#include<bits/stdc++.h>
using namespace std;

const int N = 520,INF = 1e9;
int n;
int a[N][N]; //表示每一个点
int f[N][N]; //表示状态

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ )
       for (int j = 1; j <= i; j ++ )
           cin >> a[i][j];
    
    for (int i = 0; i <= n; i ++ )
       for (int j = 0; j <= i + 1; j ++ ) //每行多初始化一个，表示三角形最右边的右上角表示负无穷
          f[i][j] = -INF; //避免处理边界情况，先将状态初始化为负无穷
    
    f[1][1] = a[1][1];
    
    for (int i = 2; i <= n; i ++ )
       for (int j = 1; j <= i; j ++ )
       f[i][j] = max(f[i - 1][j - 1] + a[i][j],f[i - 1][j] + a[i][j]);
    
    int res = -INF;   
    for (int i = 1; i <= n; i ++ ) //枚举三角形底边终点的状态的最大值
       res = max(res,f[n][i]);
       
    cout << res << endl;
           
    return 0;
}

 2.最长上升子序列

#include<bits/stdc++.h>
using namespace std;

const int N = 1010;

int n;
int a[N],f[N];

int main ()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) cin >> a[i];
    
    for (int i = 1; i <= n; i ++ )//从前往后计算每一个状态
    {
        f[i] = 1;//表示以i结尾的子序列的长度为1 只有a[i]这一个数
        for (int j = 1; j < i; j ++ )
          if(a[j] < a[i]) f[i] = max(f[i],f[j] + 1);
    }
    
    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res,f[i]); //枚举所有中点；
    
    cout << res << endl;
    
    return 0;
}

 

3.最长公共子序列

#include<bits/stdc++.h>
using namespace std;

int n,m;
const int N = 1010;
char a[N],b[N];
int f[N][N]; //状态表示为第一个字符串的前i个和第二个字符串的前j个构成的公共子序列

int main()
{
    scanf("%d %d", &n, &m);
    
    for (int i = 1; i <= n; i ++ ) cin >> a[i];
    for (int i = 1; i <= m; i ++ ) cin >> b[i];
    
    for (int i = 1; i <= n; i ++ ) 
       for (int j = 1; j <= m; j ++ )
       {
           f[i][j] = max(f[i - 1][j],f[i][j - 1]); //01和10状态包含了00状态
           //01和10分别表示a字符串的最后一个不选以及b的最后一个选
           if(a[i] == b[j]) f[i][j] = max(f[i][j],f[i - 1][j - 1] + 1);
       }
    
    cout << f[n][m] << endl;
}

 精简版：二分优化

#include<bits/stdc++.h>
using namespace std;

const int N = 100010;

int n;
int a[N]; //表示数组
int q[N]; //表示所有不同长度的上升子序列的结尾的最小值

int main ()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) cin >> a[i];
    
    int len = 0;
    q[0] = -2e9;
    
    for (int i = 0; i < n; i ++ )
    {
        int l = 0,r = len;
        while(l < r)
        {
            int mid = l + r  + 1 >> 1;
            if(q[mid] < a[i]) l = mid;
            else r = mid - 1;
        }
        len = max(len,r + 1);
        q[r + 1] = a[i];
    }
    
    cout << len << endl;
    return 0;
}

 4.最短编辑距离

#include<bits/stdc++.h>
using namespace std;

const int N = 1010;

int n,m;
char a[N],b[N];
int f[N][N]; //集合表示 由a[1 ~ i]变成b[1 ~ j]的操作方式

int main()
{
    scanf("%d%s", &n,a + 1);

    scanf("%d%s", &m,b + 1);
    
    for (int i = 0; i <= m; i ++ ) f[0][i] = i;//b的前i个字母和a的前0个字母匹配，只能将a的前i个字母增加i次
    for (int i = 0; i <= n; i ++ ) f[i][0] = i;//a的前i个字母和b的前零个字母匹配，只能将a个i个字母删除i次
    
    for (int i = 1; i <= n; i ++ )
       for (int j = 1; j <= m; j ++ )
       {
           f[i][j] = min(f[i - 1][j] + 1,f[i][j - 1] + 1);//a字符的删和增的操作
           if(a[i] == b[j]) f[i][j] = min(f[i][j],f[i - 1][j - 1]); //a字符的修改的操作
           else f[i][j] = min(f[i][j],f[i - 1][j - 1] + 1);
       }
       
    cout << f[n][m] << endl;//把a的前n个字母变成b的前m个字母
    
    return 0;
}

 5.编辑距离

#include <algorithm>
#include <cstring>
#include <iostream>

using namespace std;

const int N = 15, M = 1e3 + 10;

int n, m;
char str[M][N];
int dp[N][N];

int edit_distance(char a[], char b[])
{
    int la = strlen(a + 1), lb = strlen(b + 1);
    for (int i = 0; i <= lb; i++) {
        dp[0][i] = i;
    }
    for (int i = 0; i <= la; i++) {
        dp[i][0] = i;
    }
    for (int i = 1; i <= la; i++) {
        for (int j = 1; j <= lb; j++) {
            dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
            dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + (a[i] != b[j]));
        }
    }
    return dp[la][lb];
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> (str[i] + 1);
    }

    while (m -- ) {
        int res = 0;
        char s[N];
        int limit;
        cin >> (s + 1) >> limit;
        for (int i = 0; i < n; i ++) {
            if (edit_distance(str[i], s) <= limit) {
                res++;
            }
        }
        cout << res << endl;
    }

    return 0;
}

 

6.整数划分

 

#include<bits/stdc++.h>
using namespace std;

int n;
const int N = 1010,mod = 1e9 + 7;
int f[N];

//可视为完全背包问题去解决
int main()
{
    scanf("%d", &n);
    
    f[0] = 1;
    
    for (int i = 1; i <= n; i ++ )
       for (int j = i; j <= n; j ++ )
         f[j] = (f[j] + f[j - i]) % mod;
         
    cout << f[n] << endl;     
    
}

 第二种方式

#include<bits/stdc++.h>
using namespace std;

int n;
const int N = 1010,mod = 1e9 + 7;
int f[N][N]; //表示所有总和是i，并且恰好表示成J个数的方案

int main()
{
    scanf("%d", &n);
    
    f[0][0] = 1;
    
    for (int i = 1; i <= n; i ++ )
       for (int j = 1; j <= i; j ++ )
         f[i][j] = (f[i - 1][j - 1] + f[i - j][j]) % mod;
    
    int res = 0;     
    for (int i = 1; i <= n; i ++ ) res = (res + f[n][i]) % mod; 
    
    cout << res << endl;    
    return 0;
    
}