题目链接

题意：








题解：














AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double,double> pdd;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[8][2]={{0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};

int dp[200][10][10];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,m,d,ans=0;
    cin>>n>>m>>d;
    memset(dp,-inf,sizeof dp);
    dp[0][0][0]=0;
    for(int i=1;i<=n;i++){
        int x,y;
        cin>>x>>y;
        for(int j=d;j>=y;j--)
            for(int k=1;k<=5;k++)
                dp[j][k][0]=max(dp[j][k][0],dp[j-y][k-1][0]+x),ans=max(ans,dp[j][k][0]);
    }
    for(int i=1;i<=m;i++){
        int x,y;
        cin>>x>>y;
        for(int j=d;j>=y;j--)
            for(int k=1;k<=5;k++)
                for(int l=1;l<=k;l++)
                dp[j][k][l]=max(dp[j][k][l],dp[j-y][k][l-1]+x),ans=max(ans,dp[j][k][l]);
    }
    cout<<ans;
    return 0;
}