思路

dp(u,count)为当前再考虑下标为1-u的墙面，并且还有count免费工次的最小代价

主要是递归边界的选择：

u+1<=count return 0;
if(u==-1&&count<0)return 0x3f3f3f3f;
if(u==-1&&count==0)retrun 0;

 

这三个可以合并成

if(u<count) return 0;
if(u<0)return 0x3f3f3f3f;

 

const int N = 510;
int dp[N][2*N+10];
vector<int>times;
vector<int>costs;
int n;
int dfs(int u,int count){
    if(u<count)return 0;
    if(u==-1&&count<0)return 0x3f3f3f3f;
    if(~dp[u][count+n])return dp[u][count+n];    

    int res = 0;
    return dp[u][count+n] = min(dfs(u-1,count+times[u])+costs[u],dfs(u-1,count-1));


}

class Solution {
public:

    int paintWalls(vector<int>& cost, vector<int>& time) {
        memset(dp,-1,sizeof dp);
        costs = cost;
        times = time;
        n = cost.size();
        return dfs(n-1,0);        
    }
};