方法：Trie
本题其实就是trie的实现，每个节点需要记录两个值，深度 和 visit的次数，答案便是 max（深度 ＊ visit的次数）。
数组实现code:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'
#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);
const int maxnode = 1e7;
const int sigma_size = 2;
struct Trie
{
int ch[maxnode][sigma_size];
int val[maxnode];
int depth[maxnode];
int sz;
int ans;
Trie()
{
sz = 1;
ans = 0;
Reset(ch[0], 0);
}
void reset()
{
sz = 1;
ans = 0;
Reset(ch[0], 0);
}
void insert(string &str, int v)
{
int u = 0, n = str.length();
for (int i = 0; i < n; ++i)
{
int c = str[i] == '1';
if (!ch[u][c])
{
Reset(ch[sz], 0);
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
depth[u] = i+1;
val[u] += 1;
ans = max(ans, depth[u]*val[u]);
}
}
};
Trie root;
int main()
{
int T;
cin >> T;
cerr << T << newline;
repn(kase, T)
{
int n; cin >> n;
root.reset();
rep(i, n)
{
string str; cin >> str;
root.insert(str, 0);
}
cout << root.ans << newline;
}
}
心血来潮用动态分配空间写了一个，注意要合理释放空间，比如写一个delete（node ＊ root） 的函数，不过注意最好不要写成recursive的，容易爆栈；或者维护两个stack or queue，储存使用的node＊ 和 回收不用的node＊，code 如下
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'
#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);
struct node
{
node *child[2];
int depth, cnt;
node():depth(0), cnt(0)
{
child[0] = child[1] = nullptr;
}
};
stack<node*> ready;
stack<node*> inuse;
node* root;
node* add(int depth)
{
if (ready.empty())
{
node* temp = new node();
ready.push(temp);
}
auto ret = ready.top(); ready.pop();
ret-> depth = depth;
inuse.push(ret);
return ret;
}
int ans;
void add(const string &str)
{
auto cur = root;
int len = str.length();
for (int i = 0; i < len; ++i)
{
bool c = str[i]=='1';
if (cur->child[c] == nullptr)
{
cur->child[c] = add(i+1);
}
cur = cur->child[c];
cur->cnt += 1;
ans = max(ans, cur->cnt * cur->depth);
}
}
void clear()
{
root->child[0] = root->child[1] = nullptr;
while (!inuse.empty())
{
auto temp = inuse.top();
temp->cnt = 0;
temp->child[0] = temp->child[1] = nullptr;
ready.push(temp);
inuse.pop();
}
}
int main()
{
root = new node();
int T;
cin >> T;
cerr << T << newline;
repn(kase, T)
{
int n; cin >> n;
ans = 0;
rep(i, n)
{
string str; cin >> str; add(str);
}
cout << ans << newline;
clear();
}
}