题目就是类似背包问题，给容量为10的书架选情感值较大的书，并且这些书相邻的是不能在相同位置是相同的字母。

TLE超时了，今天是个失败的题解，求帮忙呀！

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
struct node {
    
    
    int val;
    char pre[101];
    char cur[101];
}a[2501];
bool cmp(node a, node b)
{
    
    
    if (strcmp(a.cur, b.cur) < 0)
        return 1;
    else
        return 0;
}
bool del(char* a, char* b)
{
    
       
    int i=0;
    while (a[i] != '\0' && b[i] != '\0')
    {
    
    
        if (a[i] == b[i])
            return true;
        i++;
    }
    return false;
}
int main()
{
    
    
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
    
    
        scanf("%d",&a[i].val); int sub = 0; int  sub1 = 0;
        char ch;
        ch = getchar();
        cin.getline(a[i].pre,100);
        while((ch=a[i].pre[sub++])!='\0')
        {
    
    
            if(ch!='\''&&ch!=' ')
            {
    
       
                if(ch>='A'&&ch<='Z')
                a[i].cur[sub1++]=ch+32;
                else
                a[i].cur[sub1++]=ch;
            }
        }
    }
    sort(a, a + n, cmp);
    vector<node>ans[2501][11];
    int data[2501][11] = {
    
     0 };
    for (int i = 1; i <= n; i++)
    {
    
       
        for (int j = 1; j <= 10; j++)
        {
    
    
           
            int k = i-1;
            while (ans[k][j - 1].size() != 0 && del(a[i - 1].cur, ans[k][j - 1][ans[k][j - 1].size() - 1].cur))
                {
    
       
                    k -= 1;
                    if (k == 0)
                         break;
                }
                data[i][j] = data[k ][j - 1] + a[i - 1].val;
                ans[i][j] = ans[k][j - 1];
                ans[i][j].push_back(a[i - 1]);
            if (data[i][j] < data[i - 1][j])
            {
    
    
                data[i][j] = data[i - 1][j];
                ans[i][j] = ans[i - 1][j];
            }
        }
    }
    printf("%d\n%d\n", ans[n][10].size(), data[n][10]);
    for (int i = 0; i < ans[n][10].size(); i++)
    {
    
    
        printf("%s\n", ans[n][10][i].pre);
    }
    return 0;
}