LOJ #2463. 「2018 集训队互测 Day 1」完美的旅行（目前只会循环展开大法）

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题目
枚举与值，枚举步数和最后一个点DP，枚举前一个点转移，只把起点都终点符合与值的方案计算。
最后得到的是与值为其超集的方案数，容斥一下（你也可以叫这IFMT）就可以得出答案。
时间复杂度 $O(n^3m)$
循环展开+使用 $\texttt{unsigned long long}$代替 $\texttt{long long}$ 获得 $\texttt{1000ms}$加速即可AC此题。
循环展开太强辣

AC Code:

#include <bits/stdc++.h>
#define maxn 64
#define maxm 20005
#define mod 998244353
#define LL unsigned long long
using namespace std;

int n, m, A[maxn][maxn], B[maxn][maxn], S[maxn], f[2][maxn], ans[maxn][maxm];
inline int add(int a, int b) {
    int c = a + b;
    return c >= mod ? c - mod : c;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) scanf("%d", &A[i][j]);
    for (int x = 0; x < n; x++) {
        memset(f, 0, sizeof f);
        memset(S, 0, sizeof S);
        int now = 1, pre = 0;
        for (int i = 0; i < n; i++) {
            if ((i & x) == x) {
                for (int j = 0; j < n; j++) S[j] = add(S[j], A[i][j]);
                f[pre][i] = 1;
            }
        }
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) B[i][j] = add(A[i][j], ((i & x) == x) * S[j]);
        for (int i = 0; i < n; i++) {
            LL ret = 0;
            for (int k = 0; k < n; k += 8) {
                ret += (LL)f[pre][k] * A[k][i], ret += (LL)f[pre][k + 1] * A[k + 1][i],
                    ret += (LL)f[pre][k + 2] * A[k + 2][i], ret += (LL)f[pre][k + 3] * A[k + 3][i],
                    ret += (LL)f[pre][k + 4] * A[k + 4][i], ret += (LL)f[pre][k + 5] * A[k + 5][i],
                    ret += (LL)f[pre][k + 6] * A[k + 6][i], (ret += (LL)f[pre][k + 7] * A[k + 7][i]) %= mod;
            }
            f[now][i] = ret;
            if ((i & x) == x)
                ans[x][1] = add(ans[x][1], ret);
        }
        swap(now, pre);
        for (int j = 2; j <= m; j++, swap(now, pre))
            for (int i = 0; i < n; i++) {
                LL ret = 0;
                for (int k = 0; k < n; k += 8) {
                    ret += (LL)f[pre][k] * B[k][i], ret += (LL)f[pre][k + 1] * B[k + 1][i],
                        ret += (LL)f[pre][k + 2] * B[k + 2][i], ret += (LL)f[pre][k + 3] * B[k + 3][i],
                        ret += (LL)f[pre][k + 4] * B[k + 4][i], ret += (LL)f[pre][k + 5] * B[k + 5][i],
                        ret += (LL)f[pre][k + 6] * B[k + 6][i],
                        (ret += (LL)f[pre][k + 7] * B[k + 7][i]) %= mod;
                }
                f[now][i] = ret;
                if ((i & x) == x)
                    ans[x][j] = add(ans[x][j], ret);
            }
    }
    for (int bit = 1; bit < n; bit <<= 1)
        for (int i = 0; i < n; i++)
            if (i & bit)
                for (int j = 1; j <= m; j++) ans[i - bit][j] = add(ans[i - bit][j], mod - ans[i][j]);
    int res = 0;
    for (int i = 0; i < n; i++)
        for (int j = 1; j <= m; j++) res ^= ans[i][j];
    printf("%d\n", res);
}