最长子矩阵和 删除最少的元素 回文串 最长公共上升子序列之dp复习

最长子矩阵和

#include <bits/stdc++.h>
using namespace std;
int n,ans;
int mp[105][105];
int rowAdd[105][105];
int main(){
    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> mp[i][j];
            rowAdd[i][j] = rowAdd[i][j-1] + mp[i][j];   //每一行从1到j的和
        }
    }
    for (int left = 1; left <=n ; left++) {  //左端点
        for (int right = left; right <= n; right++) {  //右端点
            int num = 0;
            for (int i = 1; i <= n; i++) {  //从第一行往下遍历，类似于前缀和的操作
                int frontAdd = rowAdd[i][right] - rowAdd[i][left-1];  //第i行第left列到第right列的总和
                if (num + frontAdd < 0) num = 0;
                else num += frontAdd;
                ans = max(ans,num);
            }
        }
    }
    cout << ans;
    return 0;
}

删除最少的元素
从前向后遍历一遍非上升子序列，再从后向前遍历一遍，dp1+dp2最大的再-1就是已经排好序的数量，用全部的数量减去这个就好了

#include <bits/stdc++.h>
using namespace std;
int n;
int arr[105];
int dp1[105];
int dp2[105];
int main(){
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> arr[i];
        dp1[i] = 1;
        dp2[i] = 1;
    }
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j < i; j++) {
            if (arr[j]>=arr[i]) dp1[i] = max(dp1[i],dp1[j]+1);
        }
    }
    for (int i = n; i >= 1; i--) {
        for (int j = n; j > i; j--) {
            if (arr[j]>=arr[i]) dp2[i] = max(dp2[i],dp2[j]+1);
        }
    }
    for (int i = 1; i <= n; i++) {
        cout << dp1[i] << " ";
    }cout << endl;
    for (int i = 1; i <= n; i++) {
        cout << dp2[i] << " ";
    }cout << endl;
    return 0;
}

回文串
reverse一下，再查最长公共子序列，得出来的长度就是已经回文的部分

最长公共上升子序列
注意dp的状态代表的意义，参考b站这位up的视频

#include <bits/stdc++.h>
using namespace std;
//最长公共上升子序列
int n,m,ans;
int arr1[105];
int arr2[105];
int dp[105][105];
int main(){
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> arr1[i];
    }
    for (int i = 1; i <= m; i++) {
        cin >> arr2[i];
    }
    //dp[i][j]表示以arr2[i]结尾的最长公共子序列
    for (int i = 1; i <= n; i++) {
        int mx = 0;
        for (int j = 1; j <= m; j++) {
            if (arr1[i]!=arr2[j]) dp[i][j] = dp[i-1][j];
            else {
                if (arr2[j-1]<arr2[j]) mx = max(1+dp[i-1][j-1],mx); //j-1位和前面字符更小且最大的dp比较
                dp[i][j] = mx;
            }
            ans = max(ans,dp[i][j]);
        }

    }
    cout << ans;
    return 0;
}