Bzoj1079:[SCOI2008]着色方案

题面

传送门

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\n", 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,\(C[i][j]\)\(C_{i}^{j}\),大小写区分开
\(f[i][j]\)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\n", f[k][0]);
    return 0;
}
posted @ 2018-01-26 09:15  Cyhlnj  阅读(...)  评论(... 编辑 收藏