BZOJ1079[SCOI2008]着色方案

設f[r1][r2][r3][r4][r5][last] 爲剩餘1 次使用機會的，剩餘2 次使用機會的…上次使用的是上次中剩餘last 的方案數

那麼記憶化搜索轉移即可

注意last 記錄的上一次中剩餘爲last 的，到本次中爲last−1 ，那麼就要減去(last==ai)

#include <cstdio>

using namespace std;

typedef long long LL;

const int SN = 17;
const int MOD = 1e9 + 7;

LL f[SN][SN][SN][SN][SN][6], n, x,rmn[SN];

void Read(LL &x) {
    LL in = 0, f = 1; char ch = getchar();
    while(ch<'0' || ch>'9') {if(ch=='-') f = -1; ch = getchar();}
    while(ch>='0' && ch<='9') {in = in*10+ch-'0'; ch = getchar();}
    x = in*f;
}

LL dfs(int r1,int r2,int r3,int r4,int r5,int last) {
    if( (r1|r2|r3|r4|r5) == 0 ) return f[r1][r2][r3][r4][r5][last] = 1;
    if(f[r1][r2][r3][r4][r5][last]) return f[r1][r2][r3][r4][r5][last];

    LL now = 0;
    if(r1) now = (now + (r1-(last==2))*dfs(r1-1,r2,r3,r4,r5,1))%MOD;
    if(r2) now = (now + (r2-(last==3))*dfs(r1+1,r2-1,r3,r4,r5,2))%MOD;
    if(r3) now = (now + (r3-(last==4))*dfs(r1,r2+1,r3-1,r4,r5,3))%MOD;
    if(r4) now = (now + (r4-(last==5))*dfs(r1,r2,r3+1,r4-1,r5,4))%MOD;
    if(r5) now = (now + r5*dfs(r1,r2,r3,r4+1,r5-1,5))%MOD;

    return f[r1][r2][r3][r4][r5][last] = now;
}

int main() {
    Read(n);

    for(int i = 1; i <= n; i++) Read(x),rmn[x]++;

    dfs(rmn[1],rmn[2],rmn[3],rmn[4],rmn[5],0);

    printf("%lld\n",f[rmn[1]][rmn[2]][rmn[3]][rmn[4]][rmn[5]][0]);

    return 0;
}