目录
题目1： P1216 [USACO1.5] [IOI1994]数字三角形 Number Triangles
题目1： P1216 [USACO1.5] [IOI1994]数字三角形 Number Triangles
代码示例：
// c++ 代码示例
#include <algorithm>
#include <iostream>
using namespace std ;
int n,a[1005][1005],f[1005][1005] ;
int dfs(int x, int y)
{
if (x == n) return a[x][y] ;
if (f[x][y] != -1) return f[x][y] ;
return f[x][y] = max(dfs(x + 1, y), dfs(x + 1, y + 1)) + a[x][y] ;
}
int main()
{
int n ;
cin >> n ;
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= i ; j++)
{
cin >> a[i][j] ;
}
}
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= i ; j++)
{
f[i][j] = -1 ;
}
}
cout << dfs(1, 1) ;
return 0 ;
}// c++ 代码示例
#include <algorithm>
#include <iostream>
using namespace std ;
long long n, a[1005][1005] ;
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 = n ; i >= 1 ; i--)
{
for (int j = 1 ; j <= i ; j++)
{
a[i][j] = a[i][j] + max(a[i + 1][j], a[i + 1][j + 1]);
}
}
cout << a[1][1] ;
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[maxn][maxn], ans;
int main()
{
read(n);
for (rint i = 1; i <= n; i++)
{
for (rint j = 1; j <= i; j++)
{
read(a[i][j]);
if (i == 1 && j == 1)
{
// The top of the triangle
continue;
}
if (j == 1)
{
// Left boundary
a[i][j] += a[i - 1][j];
}
else if (j == i)
{
// Right boundary
a[i][j] += a[i - 1][j - 1];
}
else
{
// Middle elements
a[i][j] += max(a[i - 1][j - 1], a[i - 1][j]);
}
ans = max(ans, a[i][j]);
}
}
cout << ans << endl;
return 0;
}
题目2： Common Subsequence
Common Subsequence - HDU 1159 - Virtual Judge (vjudge.net)
https://vjudge.net/problem/HDU-1159
代码示例
// c++ 代码示例
int a[MAXN], b[MAXN], f[MAXN][MAXN] ;
int dp()
{
for (int i = 1 ; i <= n ; i++)
{
for (int j = 1 ; j <= m ; j++)
{
if (a[i - 1] == b[j - 1])
{
f[i][j] = f[i - 1][j - 1] + 1 ;
}
else
{
f[i][j] = std::max(f[i - 1][j], f[i][j - 1]) ;
}
}
}
return f[n][m] ;
}// c++ 代码示例
#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
char a[1001], b[1001];
int dp[1001][1001], len1, len2;
void lcs(int i,int j)
{
for(i=1; i<=len1; i++)
{
for(j=1; j<=len2; j++)
{
if(a[i-1] == b[j-1])
dp[i][j] = dp[i-1][j-1] + 1;
else if(dp[i-1][j] > dp[i][j-1])
dp[i][j] = dp[i-1][j];
else
dp[i][j] = dp[i][j-1];
}
}
}
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[len1][len2]);
}
return 0;
}题目3 ：最长上升子序列
信息学奥赛一本通（C++版）在线评测系统 (ssoier.cn)http://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[1001] ;
int f[1001] ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a[i]) ;
f[i] = 1 ;
}
for (int i = 2 ; i <= n ; i++)
{
for (int j = i - 1 ; j >= 1 ; j--)
{
if (a[i] >= a[j])
{
f[i] = max(f[i], f[j] + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f[i]) ;
}
printf("%d", mx) ;
return 0 ;
}//c++代码示例
# include <iostream>
# include <cstdio>
using namespace std ;
int n ;
int a[1001] ;
int f[1001] ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a[i]) ;
f[i] = 1 ;
}
for (int i = n - 1 ; i >= 1 ; i--)
{
for (int j = i + 1 ; j <= n ; j++)
{
if (a[i] <= a[j])
{
f[i] = max(f[i], f[j] + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f[i]) ;
}
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[1001] ;
int f[1001] ;
int mx = -1 ;
int main()
{
scanf("%d", &n) ;
for (int i = 1 ; i <= n ; i++)
{
scanf("%d", &a[i]) ;
f[i] = 1 ;
}
for (int i = n - 1 ; i >= 1 ; i--)
{
for (int j = i + 1 ; j <= n ; j++)
{
if (a[i] < a[j])
{
f[i] = max(f[i], f[j] + 1) ;
}
}
}
for (int i = 1 ; i <= n ; i++)
{
mx = max(mx, f[i]) ;
}
printf("%d", mx) ;
return 0 ;
}
