题面
Sol1
因为每种油漆的数量是有限的
并且每种油漆是没有优先级的
直接设状态(f[lst][a][b][c][d][e])表示有(a)个可以涂一次，(b)个可以涂两次......上次涂的是可以涂(lst)次的
记搜+乘法原理即可
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(1e9 + 7);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int k, c[6];
int f[6][16][16][16][16][16];
IL int Dfs(RG int lst, RG int a, RG int b, RG int c, RG int d, RG int e){
if(f[lst][a][b][c][d][e] != -1) return f[lst][a][b][c][d][e];
if(a + b + c + d + e == 0) return 1;
RG int ret = 0, aa = a, bb = b, cc = c, dd = d, ee = e;
aa -= (lst == 2); bb -= (lst == 3); cc -= (lst == 4); dd -= (lst == 5);
if(aa > 0) (ret += 1LL * aa * Dfs(1, a - 1, b, c, d, e) % Zsy) %= Zsy;
if(bb > 0) (ret += 1LL * bb * Dfs(2, a + 1, b - 1, c, d, e) % Zsy) %= Zsy;
if(cc > 0) (ret += 1LL * cc * Dfs(3, a, b + 1, c - 1, d, e) % Zsy) %= Zsy;
if(dd > 0) (ret += 1LL * dd * Dfs(4, a, b, c + 1, d - 1, e) % Zsy) %= Zsy;
if(ee > 0) (ret += 1LL * ee * Dfs(5, a, b, c, d + 1, e - 1) % Zsy) %= Zsy;
return f[lst][a][b][c][d][e] = ret % Zsy;
}
int main(RG int argc, RG char* argv[]){
Fill(f, -1);
k = Input();
for(RG int i = 1, t; i <= k; ++i) ++c[t = Input()];
printf("%d
", Dfs(0, c[1], c[2], c[3], c[4], c[5]) % Zsy);
return 0;
}
Sol2
还有更优秀的组合数学+(DP)的做法
(sum[i])表示(c[i])的前缀和，(C[i][j])即(C_{i}^{j})，大小写区分开
设(f[i][j])表示用了前(i)种颜色涂了(sum[i])个块，其中有(j)对相邻同色块的方案数
考虑转移(f[i][j])
(c[i + 1])分成(a)组的插入到已经弄好的块中
(b)组插入到之前同色的之间
(a-b)组插空放不相邻
那么就是转移给(f[i + 1][j - b + c[i + 1] - a])
方案数为(f[i][j] * C[c[i + 1] - 1][a - 1] * C[j][b] * C[sum[i] + 1 - j][a - b])
(C[c[i + 1] - 1][a - 1])就是分成(a)组的方案数（无序的）
(C[j][b])就是插入到之前同色的之间的方案数（位置不同）
(C[sum[i] + 1 - j][a - b])就是相当于前面有(sum[i]+1)个位置，(j)个相邻块占据的位置不能放，(a-b)插入进去的方案数
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(1e9 + 7);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int k, c[16], C[80][80], f[20][80], sum[16];
IL void Prepare(){
C[0][0] = 1;
for(RG int i = 1; i <= 75; ++i){
C[i][0] = 1;
for(RG int j = 1; j <= 75; ++j)
C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Zsy;
}
}
int main(RG int argc, RG char* argv[]){
Prepare();
k = Input();
for(RG int i = 1; i <= k; ++i) c[i] = Input(), sum[i] = sum[i - 1] + c[i];
f[1][c[1] - 1] = 1;
for(RG int i = 1; i < k; ++i)
for(RG int j = 0; j < sum[i]; ++j){
if(!f[i][j]) continue;
for(RG int a = 1; a <= c[i + 1]; ++a)
for(RG int b = 0; b <= a && b <= j; ++b){
RG int ret = 1LL * f[i][j] * C[c[i + 1] - 1][a - 1] % Zsy * C[j][b] % Zsy;
ret = 1LL * ret * C[sum[i] + 1 - j][a - b] % Zsy;
(f[i + 1][j + c[i + 1] - a - b] += ret) %= Zsy;
}
}
printf("%d
", f[k][0]);
return 0;
}