基础动态规划题目基础动态规划题目
目录题目1： P1216 数字三角形 Number Triangles
代码示例：
题目2： Common Subsequence
代码示例
题目3 ：最长上升子序列
最长不下降子序列
最长上升子序列oj答案
题目1： P1216 数字三角形 Number Triangles
P1216 数字三角形 Number Triangles - 洛谷 | 盘算机科学教育新生态 (luogu.com.cn)https://csdnimg.cn/release/blog_editor_html/release2.3.6/ckeditor/plugins/CsdnLink/icons/icon-default.png?t=N7T8https://www.luogu.com.cn/problem/P1216https://i-blog.csdnimg.cn/direct/40f26514ebe6431e9a6ca12afb945afb.png
代码示例：
// c++ 代码示例
#include <algorithm>
#include <iostream>
using namespace std ;
int n,a,f ;
int dfs(int x, int y)
{
if (x == n) return a ;
if (f != -1) return f ;
return f = max(dfs(x + 1, y), dfs(x + 1, y + 1)) + a ;
}
int main()
{
int n ;
cin >> n ;
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= i ; j++)
{
cin >> a ;
}
}
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= i ; j++)
{
f = -1 ;
}
}
cout << dfs(1, 1) ;
return 0 ;
} // c++ 代码示例
#include <algorithm>
#include <iostream>
using namespace std ;
long long n, a ;
int main()
{
cin >> n ;
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= i ; j++)
{
cin >> a ;
}
}
for (int i = n ; i >= 1 ; i--)
{
for (int j = 1 ; j <= i ; j++)
{
a = a + max(a, a);
}
}
cout << a ;
return 0 ;
}
// c++ 代码示例
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
using namespace std;
#define rint register int
inline void read(int &x)
{
x = 0;
int w = 1;
char ch = getchar();
while (!isdigit(ch) && ch != '-')
{
ch = getchar();
}
if (ch == '-')
{
w = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ '0');
ch = getchar();
}
x = x * w;
}
const int maxn = 1000 + 10;
int n, a, ans;
int main()
{
read(n);
for (rint i = 1; i <= n; i++)
{
for (rint j = 1; j <= i; j++)
{
read(a);
if (i == 1 && j == 1)
{
// The top of the triangle
continue;
}
if (j == 1)
{
// Left boundary
a += a;
}
else if (j == i)
{
// Right boundary
a += a;
}
else
{
// Middle elements
a += max(a, a);
}
ans = max(ans, a);
}
}
cout << ans << endl;
return 0;
} https://i-blog.csdnimg.cn/direct/143ab111e9ca4e7b9251b0aeeeae0f7c.png
题目2： Common Subsequence
Common Subsequence - HDU 1159 - Virtual Judge (vjudge.net)https://csdnimg.cn/release/blog_editor_html/release2.3.6/ckeditor/plugins/CsdnLink/icons/icon-default.png?t=N7T8https://vjudge.net/problem/HDU-1159
代码示例
// c++ 代码示例
int a, b, f ;
int dp()
{
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= m ; j++)
{
if (a == b)
{
f = f + 1 ;
}
else
{
f = std::max(f, f) ;
}
}
}
return f ;
} // c++ 代码示例
#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
char a, b;
int dp, len1, len2;
void lcs(int i,int j)
{
for(i=1; i<=len1; i++)
{
for(j=1; j<=len2; j++)
{
if(a == b)
dp = dp + 1;
else if(dp > dp)
dp = dp;
else
dp = dp;
}
}
}
int main()
{
while(~scanf(" %s",a))
{
scanf(" %s", b);
memset(dp, 0, sizeof(dp));
len1 = strlen(a);
len2 = strlen(b);
lcs(len1, len2);
printf("%d\n", dp);
}
return 0;
} 题目3 ：最长上升子序列
信息学奥赛一本通（C++版）在线评测体系 (ssoier.cn)https://csdnimg.cn/release/blog_editor_html/release2.3.6/ckeditor/plugins/CsdnLink/icons/icon-default.png?t=N7T8http://ybt.ssoier.cn:8088/problem_show.php?pid=1281
最长不下降子序列
//c++代码示例
# include <iostream>
# include <cstdio>
using namespace std ;
int n ;
int a ;
int f ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a) ;
f = 1 ;
}
for (int i = 2 ; i <= n ; i++)
{
for (int j = i - 1 ; j >= 1 ; j--)
{
if (a >= a)
{
f = max(f, f + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f) ;
}
printf("%d", mx) ;
return 0 ;
} //c++代码示例
# include <iostream>
# include <cstdio>
using namespace std ;
int n ;
int a ;
int f ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a) ;
f = 1 ;
}
for (int i = n - 1 ; i >= 1 ; i--)
{
for (int j = i + 1 ; j <= n ; j++)
{
if (a <= a)
{
f = max(f, f + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f) ;
}
printf("%d", mx) ;
return 0 ;
} 最长上升子序列oj答案
//c++代码示例
# include <iostream>
# include <cstdio>
using namespace std ;
int n ;
int a ;
int f ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a) ;
f = 1 ;
}
for (int i = n - 1 ; i >= 1 ; i--)
{
for (int j = i + 1 ; j <= n ; j++)
{
if (a < a)
{
f = max(f, f + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f) ;
}
printf("%d", mx) ;
return 0 ;
}
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