[NOI.AC#35]string 缩点+拓扑排序

链接

因为有交换相邻字母，因此给你字符串就相当于给你了这个字符串的所有排列

把等价的串映射到整数范围，再根据 \(m\) 种魔法连边，缩点后在 DAG 上DP即可

无耻地用了int128

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i(a);i<=(b);++i)
#define dbg(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef __int128 ll;
typedef unsigned int uint;
typedef unsigned long long ull;
template<typename T,typename U>inline char smin(T&x,const U&y){return x>y?x=y,1:0;}
template<typename T,typename U>inline char smax(T&x,const U&y){return x<y?x=y,1:0;}
const int N=52,MN=N*N*N;
int n,m,c1[N<<1][4],c2[N<<1][4],num[N][N][N],T;
ll C[N][N],cnt[MN],val[MN],f[MN],ans;
char s1[N],s2[N];
int head[MN],fr[MN*N],to[MN*N],nxt[MN*N],tot;
inline void add(int x,int y){if(x==y)return;fr[++tot]=x,to[tot]=y,nxt[tot]=head[x];head[x]=tot;}
int s[MN],top,dfn[MN],dfs_clock,bel[MN],scc;
bool ins[MN];
int dfs(int x){
   int low=dfn[x]=++dfs_clock;
   s[++top]=x,ins[x]=1;
   for(int i=head[x];i;i=nxt[i]){
       const int&y=to[i];
       if(!dfn[y])smin(low,dfs(y));
       else if(ins[y])smin(low,dfn[y]);
   }
   if(low==dfn[x]){
       ++scc;
       do bel[s[top]]=scc,val[scc]+=cnt[s[top]],ins[s[top]]=0;while(s[top--]!=x);
   }
   return low;
}
int ind[MN],q[MN];
void toposort(){
   int l=1,r=0,x;
   REP(i,1,scc)if(!ind[i])q[++r]=i;
   while(l<=r){
       x=q[l++];f[x]+=val[x];smax(ans,f[x]);
       for(int i=head[x];i;i=nxt[i]){
           smax(f[to[i]],f[x]);
           if(!--ind[to[i]])q[++r]=to[i];
       }
   }
}
template<typename Tp>inline void print(Tp x){
   static char s[50];
   int top=0;
   for(;x;x/=10)s[++top]=x%10^'0';
   while(top)putchar(s[top--]);
}
int main(){
   scanf("%d%d",&n,&m);
   REP(i,1,m){
       scanf("%s%s",s1+1,s2+1);int len=strlen(s1+1);
       REP(j,1,len)++c1[i][s1[j]-'A'],++c2[i][s2[j]-'A'];
   }
   C[0][0]=1;
   REP(i,1,n){
       C[i][0]=1;
       REP(j,1,i)C[i][j]=C[i-1][j]+C[i-1][j-1];
   }
   REP(i,0,n)REP(j,0,n-i)REP(k,0,n-i-j){
       num[i][j][k]=++T;
       cnt[T]=C[n][i]*C[n-i][j]*C[n-i-j][k];
   }
   REP(a,0,n)REP(b,0,n-a)REP(c,0,n-a-b){
       int d=n-a-b-c;
       REP(i,1,m)if(a>=c1[i][0]&&b>=c1[i][1]&&c>=c1[i][2]&&d>=c1[i][3]){
           int e,f,g,h;
           e=a-c1[i][0]+c2[i][0];
           f=b-c1[i][1]+c2[i][1];
           g=c-c1[i][2]+c2[i][2];
           add(num[a][b][c],num[e][f][g]);
       }
   }
   REP(i,1,T)if(!dfn[i])dfs(i);
   int all=tot;tot=0;memset(head,0,sizeof head);
   REP(i,1,all){
       int&bx=bel[fr[i]],&by=bel[to[i]];
       if(bx!=by)add(bx,by),++ind[by];
   }
   toposort();print(ans);
   return 0;
}

转载于:https://www.cnblogs.com/HolyK/p/9839758.html