思路

有点麻烦的Trie树题

注意到每个节点的贡献是(深度（根的深度是0）乘2+1)乘上在这个点分叉的字符串个数，所以全部插入之后dfs一遍即可（为了避免两个字符串一样的情况，还应该特判一下最后的中止节点）

代码

字符集太大需要用邻接表存才能跑过

直接存边的TLE代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#define int long long
using namespace std;
signed Trie[4000100][65];
int Nodecnt,sz[4000100],ans=0,root,n;
char s[4010][1010];
int tonum(char c){if(c>='a'&&c<='z')return c-'a';else if(c=='a'+26)return 26;else if(c>='A'&&c<='Z')return c-'A'+27;else if(c>='0'&&c<='9')return c-'0'+54;
}
void insert(char *s,int len){int o=root;sz[o]++;for(int i=0;i<len;i++){if(!Trie[o][tonum(s[i])])Trie[o][tonum(s[i])]=++Nodecnt;o=Trie[o][tonum(s[i])];sz[o]++;}
}
void dfs(int o,int dep,int isend=0){if(!isend){int midt=0;for(int i=0;i<65;i++){if(Trie[o][i]){sz[o]-=sz[Trie[o][i]];midt+=sz[Trie[o][i]]*sz[o];}}ans+=(dep+1)*midt;for(int i=0;i<65;i++){if(Trie[o][i]){dfs(Trie[o][i],dep+2,(i==26));}}}else{ans+=(sz[o]-1)*(sz[o])/2*dep;}
}
void init(void){for(int i=0;i<=Nodecnt;i++){for(int j=0;j<65;j++)Trie[i][j]=0;sz[i]=0;}ans=0;Nodecnt=1;root=1;
}
signed main(){// freopen("test.in","r",stdin);// freopen("test.out","w",stdout);Nodecnt=1;root=1;ans=0;int cnt=0;while(scanf("%lld",&n)==1&&n){++cnt;for(int i=1;i<=n;i++){scanf("%s",s[i]+1);int len=strlen(s[i]+1);s[i][len+1]='a'+26;insert(s[i]+1,len+1);}dfs(root,0);printf("Case %lld: %lld\n",cnt,ans);init();}return 0;
}

邻接表存边的AC代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#define int long long
using namespace std;
signed fir[4000100],nxt[4000100],v[4000100],num[4000100],Ecnt=0;
int Nodecnt,sz[4000100],ans=0,root,n;
char s[4010][1010];
int tonum(char c){if(c>='a'&&c<='z')return c-'a';else if(c=='a'+26)return 26;else if(c>='A'&&c<='Z')return c-'A'+27;else if(c>='0'&&c<='9')return c-'0'+54;
}
void addedge(int ui,int vi){++Ecnt;v[Ecnt]=vi;nxt[Ecnt]=fir[ui];fir[ui]=Ecnt;
}
void insert(char *s,int len){int o=root;sz[o]++;for(int i=0;i<len;i++){int vx;bool f=false;for(int j=fir[o];j;j=nxt[j])if(num[v[j]]==tonum(s[i])){vx=v[j];f=true;break;}if(!f){vx=++Nodecnt;num[vx]=tonum(s[i]);addedge(o,vx);}o=vx;sz[o]++;}
}
void dfs(int o,int dep,int isend=0){if(!isend){int midt=0;for(int i=fir[o];i;i=nxt[i]){sz[o]-=sz[v[i]];midt+=sz[v[i]]*sz[o];}ans+=(dep+1)*midt;for(int i=fir[o];i;i=nxt[i]){dfs(v[i],dep+2,(num[v[i]]==26));}}else{ans+=(sz[o]-1)*(sz[o])/2*dep;}
}
void init(void){ans=0;Nodecnt=1;root=1;Ecnt=0;memset(sz,0,sizeof(sz));memset(fir,0,sizeof(fir));memset(nxt,0,sizeof(nxt));memset(v,0,sizeof(v));memset(num,0,sizeof(num));
}
signed main(){// freopen("test.in","r",stdin);// freopen("test.out","w",stdout);Nodecnt=1;root=1;ans=0;int cnt=0;while(scanf("%lld",&n)==1&&n){++cnt;for(int i=1;i<=n;i++){scanf("%s",s[i]+1);int len=strlen(s[i]+1);s[i][len+1]='a'+26;insert(s[i]+1,len+1);}dfs(root,0);printf("Case %lld: %lld\n",cnt,ans);init();}return 0;
}