[dp+数论]bzoj1004

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具体解法关键的就两句话

每个置换的核染色方案数相加/置换数=总染色方案数

每个置换的核染色方案数=缩环后的染色方案数

#include<iostream>#include<cstdlib>#include<cmath>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int f[31][31][31];int cg[71];bool flag[71];int sum[71],id[71],st[71];int dfs(int x){    if (flag[x]) return 0;    flag[x]=true;    return dfs(cg[x])+1;}int r,b,g,m,p,n,P;int main(){    freopen("input.txt","r",stdin);    long long ans=0;    scanf("%d%d%d%d%d",&r,&b,&g,&m,&P);    n=r+b+g;    m++;    for (int i=1;i<=m;i++)    {        if (i==m) for (int j=1;j<=n;j++) cg[j]=j;        else for (int j=1;j<=n;j++) scanf("%d",&cg[j]);        memset(st,0,sizeof(st));        memset(id,0,sizeof(id));        memset(sum,0,sizeof(sum));        memset(f,0,sizeof(f));        memset(flag,0,sizeof(flag));        for (int j=1;j<=n;j++)            if (!flag[j])                st[++st[0]]=dfs(j);        for (int j=1;j<=st[0];j++)        {            sum[j]=sum[j-1]+st[i];            id[sum[j]]=j;        }        f[0][0][0]=1;        int p;        for (int rr=0;rr<=r;rr++)            for (int bb=0;bb<=b;bb++)                for (int gg=0;gg<=g;gg++)                    if (p=id[rr+bb+gg])                    {                        if (rr>=st[p]) f[rr][bb][gg]=(f[rr][bb][gg]+f[rr-st[p]][bb][gg])%P;                        if (bb>=st[p]) f[rr][bb][gg]=(f[rr][bb][gg]+f[rr][bb-st[p]][gg])%P;                        if (gg>=st[p]) f[rr][bb][gg]=(f[rr][bb][gg]+f[rr][bb][gg-st[p]])%P;                    }        ans=(ans+f[r][b][g])%P;    }    int now=1;    for (int i=P-2;i;i=i>>1)    {        if (i&1) now=((long long)now*m)%P;        m=((long long)m*m)%P;    }    printf("%lld",(ans*now)%P);    return 0;}

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