2021台北ICPC Seesaw (背包)

• 问题转化为从A，B两组中分别取相同个数的东西的某一权值总和的最取次数
• fa/b[i][j][v] 表示前i个数，取j个，能否凑成体积v，f值就是0/1
• 注意控制合法状态，最后的枚举不要超过A，B的值域

#include<bits/stdc++.h>
//#include <bits/extc++.h>
using namespace std;
// using namespace __gnu_cxx;
// using namespace __gnu_pbds;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);  
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
#define li __int128_t
//#define int long long
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int mod = 1e9 + 7;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
int a[105], b[105], cnta, cntb;
bool fa[105][105][5055], fb[105][105][5055];
void solve() {
    cnta = cntb = 0;
    memset(fa, 0, sizeof fa);
    memset(fb, 0, sizeof fb);
    int n, x; cin >> n;
    int suma = 0, sumb = 0, A = 0, B = 0;
    for ( int i = 1; i <= n; ++ i ) {
        cin >> x; 
        if( x <= 2 ) b[ ++ cntb ] = i, B += i;
        sumb += x * i;
    }
    for ( int i = 1; i <= n; ++ i ) {
        cin >> x; 
        if( x <= 2 ) a[ ++ cnta ] = i, A += i;
        suma += x * i;
    }
    if( suma == sumb ) {
        cout << 0 << '\n'; return;
    }
    if( suma < sumb) {
        cout << -1 << '\n'; return;
    }
    for ( int i = 0; i <= cnta; ++ i) fa[i][0][0] = 1;
    for ( int i = 0; i <= cntb; ++ i) fb[i][0][0] = 1; //第二位是要控制合法状态，第一维不用
    for ( int i = 1; i <= cnta; ++ i ) {
        for ( int j = 1; j <= cnta; ++ j ) {
            for ( int v = A; v >= 0; -- v ) {//要继承
                if( v >= a[i] && fa[i - 1][j - 1][v - a[i]] )
                fa[i][j][v] = fa[i - 1][j - 1][v - a[i]];
                if( fa[i - 1][j][v] > 0)
                fa[i][j][v] = fa[i - 1][j][v];//要继承
            }
        }
    }

    for ( int i = 1; i <= cntb; ++ i ) {
        for ( int j = 1; j <= cntb; ++ j ) {
            for ( int v = B; v >= 0; -- v ) { //要继承
                if( v >= b[i] && fb[i - 1][j - 1][v - b[i]] )
                fb[i][j][v] = fb[i - 1][j - 1][v - b[i]];
                if( fb[i - 1][j][v] > 0)
                fb[i][j][v] = fb[i - 1][j][v]; //要继承
            }
        }
    }
    int cha = suma - sumb;
    for ( int i = 1; i <= min(cnta, cntb); ++ i ) {
        int st = min(A, cha); //要在A，B值域内枚举
        for ( int j = st; j >= 0; -- j ) {
            if(cha - j > B) continue; // cha - j 会特别大，就越界了
            if(fa[cnta][i][j] > 0 && fb[cntb][i][cha - j] > 0 ) {
                cout << i << '\n';
                return;
            }
        }
    }
    cout << -1 <<'\n';
}
int main() {
    IOS
    int t; cin >> t; 
    while( t -- ) solve();
    return 0;
}