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loj#10172 涂抹果酱 (状压DP)

题目:

#10172. 「一本通 5.4 练习 1」涂抹果酱

解析:

三进制的状压DP
经过简单的打表发现,在\(m=5\)时最多有\(48\)种合法状态
然后就向二进制一样枚举当前状态和上一层的状态进行转移就好了
由于第\(k\)行是给定的,所以转移时要特判一下第\(k\)行,并且注意下一\(k=1\)的情况
\(f[i][j]\)表示第\(i\)\(j\)状态时的方案数
\(f[i][j] += f[i - 1][k],k是上一行的状态\)

代码:

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e5 + 10;
const int mod = 1e6;

int n, m, k, num, sum = 1, sta, kk, ans;
/*
    sum为状态上节
    sta是第k行状态
    num是合法状态数
*/
int state[N], f[10010][50];
//state记录状态
//f[i][j]表示dp 到第i行j状态的方案数

bool check(int x) {     //判断状态是否合法
    int len = 0;
    int b[10];
    while (x) b[++len] = x % 3, x /= 3;
    if (len < m - 1) return 0;
    for (int i = 2; i <= len; ++i) if (b[i] == b[i - 1]) return 0;
    return 1;   
}

bool cmp(int x, int y) {    //判断两个状态是否可以相邻
    for (int i = 1; i <= m; ++i) {
        if ((x % 3) == (y % 3)) return 0;
        x /= 3, y /= 3;
    }
    return 1;
}

signed main() {
    cin >> n >> m >> k;
    for (int i = 1, x; i <= m; ++i) cin >> x, kk = kk * 3 + x - 1;
    for (int i = 1; i <= m; ++i) sum *= 3;
    for (int i = 0; i < sum; ++i) if (check(i)) {
            state[++num] = i;
            if (i == kk) sta = num;
        }
    if (sta == 0) {
        cout << 0;
        return 0;
    }
    for (int i = 1; i <= n; ++i) {
        if (i == k) {
            if (i == 1) f[1][sta] = 1;
            else for (int j = 1; j <= num; ++j)
                    if (cmp(state[sta], state[j]))
                        (f[i][sta] += f[i - 1][j]) %= mod;
        } else {
            if (i == 1) for (int j = 1; j <= num; ++j) f[i][j] = 1;
            else for (int j = 1; j <= num; ++j)
                    for (int pre = 1; pre <= num; ++pre)
                        if (cmp(state[j], state[pre])) 
                            (f[i][j] += f[i - 1][pre]) %= mod;
        }
    }
    for (int i = 1; i <= num; ++i) (ans += f[n][i]) %= mod;
    cout << ans;
}
posted @ 2019-10-10 19:59 Chrety 阅读(...) 评论(...) 编辑 收藏