传送门:HDU - 6804
思路分析
将B组的重量改为相反数,问题就转化为01背包问题,答案是重量为0的最大价值
因为值域可能会很大,考虑将数组随机,值域就会减小
AC代码
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int base=1e6;
const int ch=150000;
ll dp[base+ch*3];
struct node{
int w,v;
}a[3000];
#define re register
int main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int T;
read(T);
while(T--){
int n,m;
read(n,m);
for(re int i=1;i<=n+m;++i){
read(a[i].v,a[i].w);
if(i>n) a[i].v=-a[i].v;
}
n+=m;
for(re int i=base-ch;i<=base+ch;++i) dp[i]=-1e18;
random_shuffle(a+1,a+1+n);
random_shuffle(a+1,a+1+n);
dp[base]=0;
for(re int i=1;i<=n;i++){
if(a[i].v>=0){
for(int j=base+ch;j>=base-ch+a[i].v;j--) dp[j]=max(dp[j],dp[j-a[i].v]+a[i].w);
}else{
for(int j=base-ch;j<=base+ch-a[i].v;j++) dp[j]=max(dp[j],dp[j-a[i].v]+a[i].w);
}
}
printf("%lld\n",dp[base]);
}
return 0;
}