附上原文地址
LCS:https://www.cnblogs.com/GetcharZp/p/9070367.html
for(int i = 1; i <= len1; ++ i)
{
for(int j = 1; j <= len2; ++ j)
{
if(s1[i] == s2[j]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
}
}
LIS、LDS:https://blog.csdn.net/qq_37753409/article/details/78665536
const int MAXN = 100005;
int a[MAXN], dp[MAXN];
//最长上升子序列
int LIS(int n)
{
int res = 0;
for(int i = 0; i < n; ++i)
{
dp[i] = 1;
for(int j = 0; j < i; ++j)
if(a[j] < a[i]) dp[i] = max(dp[i], dp[j]+1);
res = max(res, dp[i]);
}
return res;
}
//最长下降子序列
int LDS(int n)
{
int res = 0;
for(int i = 0; i < n; ++i)
{
dp[i] = 1;
for(int j = 0; j < i; ++j)
if(a[j] > a[i]) dp[i] = max(dp[i], dp[j] + 1);
res = max(res, dp[i]);
}
return res;
}