卡常题解

题解如下：

#include <iostream> 
#include <vector> 
#pragma GCC optimize("Ofast") 
#pragma GCC optimize(3) 
using namespace std; 
bool f(const vector<vector<long long>>& d, const vector<long long>& n, long long m, const vector<long long>& t) { 
    for (size_t k = 0; k < n.size();  ++k) { 
        if (t[k] < n[k]) { 
            return false; 
        } 
    } 
    return true; 
} 
bool g(vector<vector<long long>>& d, vector<long long>& n, long long x, long long y) { 
    vector<bool> dp(1 << x, false); 
    dp[0]=true; 
    vector<long long> t(y, 0); 
    for (long long i = 0; i < x; ++i) { 
        for (long long j = 0; j < y; ++j) { 
            t[j]+=d[i][j]; 
        } 
    } 
    if (f(d, n, 0, t)) { 
        return true; 
    } 
    for (long long m = 1; m < (1 << x); ++m) { 
        vector<long long> t(y, 0); 
        for (long long i = 0; i < x; ++i) { 
            if (m & (1 << i)) { 
                for (long long j = 0; j < y; ++j) { 
                    t[j]+=d[i][j]; 
                } 
            } 
        } 
        dp[m]=false; 
        for (long long subm = m; subm > 0; subm = (subm - 1)&m) { 
            if (dp[subm] && f(d, n, subm, t)) { 
                dp[m]=true; 
                break; 
            } 
        } 
    } 
    return dp[(1 << x)-1]; 
}  
int main() { 
    cin.tie(nullptr);  
    ios::sync_with_stdio(0); 
    long long x, y; 
    cin >> x >> y; 
    vector<long long> n(y); 
    for (long long i = 0; i < y; ++i) { 
        cin >> n[i]; 
    } 
    vector<vector<long long>> d(x, vector<long long>(y)); 
    for (long long i = 0; i < x; ++i) { 
        for (long long j = 0; j < y; ++j) { 
            cin >> d[i][j]; 
        } 
    } 
    if (g(d, n, x, y)) { 
        cout << "Yes" << endl; 
    } else { 
        cout << "No" << endl; 
    } 
    return 0; 
}