2020杭电多校联合训练（第四场） C.Contest of Rope Pulling （01背包指令集优化）

题面

Problem Description
Rope Pulling, also known as Tug of War, is a classic game. Zhang3 organized a rope pulling contest between Class 1 and Class 2.

There are n students in Class 1 and m students in Class 2. The ith student has strength wi and beauty-value vi. Zhang3 needs to choose some students from both classes, and let those chosen from Class 1 compete against those chosen from Class 2. It is also allowed to choose no students from a class or to choose all of them.

To be a fair contest, the total strength of both teams must be equal. To make the contest more beautiful, Zhang3 wants to choose such a set of students, that the total beauty-value of all participants is maximized. Please help her determine the optimal total beauty-value.

Input
The first line of the input gives the number of test cases, T(1≤T≤30). T test cases follow.

For each test case, the first line contains two integers n,m(1≤n,m≤1000), representing the number of students in Class 1 and Class 2.

Then (n+m) lines follow, describing the students. The ith line contains two integers wi,vi(1≤wi≤1000,−109≤vi≤109), representing the strength and the beauty-value of the ith student. The first n students come from Class 1, while the other m students come from Class 2.

The sum of (n+m) in all test cases doesn't exceed 104.

Output
For each test case, print a line with an integer, representing the optimal total beauty-value.

Sample Input
2
3 4
4 7
3 8
2 2
1 4
5 8
1 3
4 4
1 2
1000 -10000
200 3000
800 5000

Sample Output
30
0

思路

暴力跑背包一定是会超时的，所以我们考虑指令集优化，复杂度n3.正解的话，呶，看下面，我看不懂...

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=0;i<end;i++)
#define inf 0x3f3f3f3f3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f = -1;
        ch=getchar();
    } 
    while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }   return x*f;
}
const int maxn=1e6+10;
ll dp[maxn];
ll dp2[maxn];
int t;
int n,m;

void solve () {
    int sum1=0,sum2=0;
    rep (i,1,n) {
        int x,y;
        cin>>x>>y;
        per (j,sum1,0) {
            if (dp[j]+y>dp[j+x]) dp[j+x]=dp[j]+y;
        }
        sum1+=x;
    }
    rep (i,1,m) {
        int x,y;
        cin>>x>>y;
        per (j,sum2,0) {
            if (dp2[j]+y>dp2[j+x]) dp2[j+x]=dp2[j]+y;
        }
        sum2+=x;
    }
    ll ans=0;
    int sum=min (sum1,sum2);
    rep (i,1,sum) {
        ans=max (ans,dp[i]+dp2[i]);
    }
    cout<<ans<<endl;
    rep (i,1,sum2) dp[i]=-1e15;
    rep (i,1,sum2) dp2[i]=-1e15; 
}

int main () {
   cin>>t;
   rep (i,1,maxn) {
       dp[i]=dp2[i]=-1e15;
   }
   while (t--) {
       cin>>n>>m;
       solve ();
   }
    return 0;
}