链接
思路
设 f [ i ] [ j ] f[i][j] f[i][j] 表示体积为 j j j 的情况下且异或和为 i i i 的方案是否存在。
使用 b i t s e t bitset bitset 优化。
状态转移方程:
f[j] |= g[j ^ w[i]] << v[i]
g[j]
表示上一层的状态。
代码
// #pragma GCC optimize(3)
#include <set>
#include <map>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <cstdio>
#include <bitset>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#define endl '\n'
#define fi first
#define se second
#define PI acos(-1)
// #define PI 3.1415926
#define LL long long
#define INF 0x3f3f3f3f
#define lowbit(x) (-x&x)
#define PII pair<int, int>
#define ULL unsigned long long
#define PIL pair<int, long long>
#define mem(a, b) memset(a, b, sizeof a)
#define rev(x) reverse(x.begin(), x.end())
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
const int N = 1050;
int n, m;
int v[N], w[N];
void solve() {
int tt;
cin >> tt;
while (tt -- ) {
bitset<N> f[N], g[N];
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];
f[0][0] = 1;
for (int i = 1; i <= n; i ++ ) {
for (int j = 0; j < 1024; j ++ ) g[j] = f[j] << v[i];
for (int j = 0; j < 1024; j ++ ) f[j] |= g[j ^ w[i]];
}
int ans = -1;
for (int i = 1023; i >= 0; i -- ) {
if (f[i][m]) {
ans = i;
break;
}
}
cout << ans << endl;
}
}
int main() {
IOS;
solve();
return 0;
}