CodeCraft-19 and Codeforces Round #537 (Div. 2) - D. Destroy the Colony（动态规划+组合数学）

Problem CodeCraft-19 and Codeforces Round #537 (Div. 2) - D. Destroy the Colony

Time Limit: 2000 mSec

Problem Description

Input

Output

For each question output the number of arrangements possible modulo

Sample Input

abba
2
1 4
1 2

Sample Output

2
0

题解：这个题的每一部分都比较简单，加在一起就有难度了。关于组合数学的部分就不说了，比较简单，这个题的精华在于dp的部分，对于每次的询问，相当于强制让两种字符不在左边（或者不在右边），也就是在原始的dp递推方程中要跳过这两个字符的方案数，如果每次都重新递推，递推的复杂度是O(n * k)，一共有k^2个组合，复杂度会炸，这时就要考虑利用之前dp的信息来减少递推的计算量，新的tmp_dp[j]的状态定义长度为j的，没有第i种字符的方案数，这其实就是长度为j的总方案数减去强制有第i种字符的方案数，而强制有i的方案数就应该是没有i的长度为j - cnt[i]的方案数，这恰好就是tmp_dp[j - cnt[i]]，因此tmp[j] = dp[j] - tmp_dp[j - cnt[i]]，可以在O(n)的时间内完成递推，一共有k^2个组合，因此复杂度时O(n * k^2)，可以接受。

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 #define REP(i, n) for (int i = 1; i <= (n); i++)
  6 #define sqr(x) ((x) * (x))
  7 
  8 const int maxn = 100000 + 10;
  9 const int maxm = 200000 + 100;
 10 const int maxs = 52;
 11 
 12 typedef long long LL;
 13 typedef pair<int, int> pii;
 14 typedef pair<double, double> pdd;
 15 
 16 const LL unit = 1LL;
 17 const int INF = 0x3f3f3f3f;
 18 const LL Inf = 0x3f3f3f3f3f3f3f3f;
 19 const double eps = 1e-14;
 20 const double inf = 1e15;
 21 const double pi = acos(-1.0);
 22 const LL mod = 1000000007;
 23 
 24 inline int cal(char ch)
 25 {
 26     if (isupper(ch))
 27         return ch - 'A' + 26;
 28     return ch - 'a';
 29 }
 30 
 31 LL pow_mod(LL x, LL n, LL mod)
 32 {
 33     LL ans = 1, base = x;
 34     while (n)
 35     {
 36         if (n & 1)
 37             ans = (ans * base) % mod;
 38         base = (base * base) % mod;
 39         n >>= 1;
 40     }
 41     return ans;
 42 }
 43 
 44 void Add(LL &a, LL b)
 45 {
 46     a += b;
 47     if (a > mod)
 48         a -= mod;
 49 }
 50 
 51 void Sub(LL &a, LL b)
 52 {
 53     a -= b;
 54     if (a < 0)
 55         a += mod;
 56 }
 57 
 58 string str;
 59 LL n;
 60 LL cnt[maxs];
 61 LL fact[maxn], inv[maxn];
 62 LL dp[maxn], tmp_dp[maxn];
 63 LL ans[maxs][maxs];
 64 
 65 int main()
 66 {
 67     ios::sync_with_stdio(false);
 68     cin.tie(0);
 69     //freopen("input.txt", "r", stdin);
 70     //freopen("output.txt", "w", stdout);
 71     cin >> str;
 72     n = str.size();
 73     for (int i = 0; i < n; i++)
 74     {
 75         cnt[cal(str[i])]++;
 76     }
 77     fact[0] = fact[1] = 1;
 78     for (LL i = 2; i <= n; i++)
 79     {
 80         fact[i] = (fact[i - 1] * i) % mod;
 81     }
 82     inv[n] = pow_mod(fact[n], mod - 2, mod);
 83     for (LL i = n - 1; i >= 0; i--)
 84     {
 85         inv[i] = (inv[i + 1] * (i + 1)) % mod;
 86     }
 87     LL num = (fact[n / 2] * fact[n / 2]) % mod;
 88     for (int i = 0; i < maxs; i++)
 89     {
 90         num = (num * inv[cnt[i]]) % mod;
 91     }
 92     dp[0] = 1;
 93     for (int i = 0; i < maxs; i++)
 94     {
 95         if (!cnt[i])
 96             continue;
 97         for (int j = n; j >= cnt[i]; j--)
 98         {
 99             Add(dp[j], dp[j - cnt[i]]);
100         }
101     }
102     for (int i = 0; i < maxs; i++)
103     {
104         ans[i][i] = dp[n / 2];
105     }
106     for (int i = 0; i < maxs; i++)
107     {
108         if (!cnt[i])
109             continue;
110         for (int i = 0; i <= n; i++)
111             tmp_dp[i] = dp[i];
112         for (int t = cnt[i]; t <= n; t++)
113         {
114             Sub(tmp_dp[t], tmp_dp[t - cnt[i]]);
115         }
116         for (int j = i + 1; j < maxs; j++)
117         {
118             if (!cnt[j])
119                 continue;
120             for (int t = cnt[j]; t <= n; t++)
121             {
122                 Sub(tmp_dp[t], tmp_dp[t - cnt[j]]);
123                 ans[i][j] = (2LL * tmp_dp[n / 2]) % mod;
124                 ans[j][i] = ans[i][j];
125             }
126             for (int t = n; t >= cnt[j]; t--)
127                 Add(tmp_dp[t], tmp_dp[t - cnt[j]]);
128         }
129     }
130     int q;
131     cin >> q;
132     int x, y;
133     while (q--)
134     {
135         cin >> x >> y;
136         int l = cal(str[x - 1]);
137         int r = cal(str[y - 1]);
138         cout << (num * ans[l][r]) % mod << "\n";
139     }
140     return 0;
141 }