Codeforces 279D The Minimum Number of Variables 状压dp

我们定义dp[ i ][ mask ]表示是否存在处理完前 i 个a， b中存者 a存在的状态是mask 的情况。

然后用sosdp处理出，状态为state的a，能组成的数字，然后转移就好啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n;
int a[N];

int cnt[1 << 23];
int sos[1 << 23];
bool dp[23][1 << 23];

int main() {
    scanf("%d", &n);
    for(int i = 1; i < (1 << n); i++)
        cnt[i] = cnt[i - (i & - i)] + 1;
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < n; i++) {
        for(int j = i; j < n; j++) {
             int ret = a[i] + a[j];
             for(int k = 0; k < n; k++) {
                if(ret == a[k])
                    sos[(1 << i) | (1 << j)] |= 1 << k;
             }
        }
    }
    for(int i = 0; i < n; i++)
        for(int mask = 0; mask < (1 << n); mask++)
            if(mask >> i & 1) sos[mask] |= sos[mask ^ (1 << i)];

    dp[0][1] = 1;
    for(int i = 0; i < n - 1; i++) {
        for(int mask = 0; mask < (1 << n); mask++) {
            if(!dp[i][mask]) continue;
            if(sos[mask] >> (i + 1) & 1) {
                dp[i + 1][mask | (1 << (i + 1))] = true;
                for(int j = 0; j <= i; j++) {
                    if(mask >> j & 1) {
                        dp[i + 1][(mask ^ (1 << j)) | (1 << (i + 1))] = true;
                    }
                }
            }
        }
    }
    int ans = inf;
    for(int i = 1; i < (1 << n); i++)
        if(dp[n - 1][i]) chkmin(ans, cnt[i]);
    if(ans == inf) ans = -1;
    printf("%d\n", ans);
    return 0;
}

/*
*/

Codeforces 279D The Minimum Number of Variables 状压dp

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