题目大意
给出一个残缺的字符串，每个位置都 (in[0,9])。有 (m) 中贡献，即 (s,k)，表示该字符串中没出现一次 (s)，贡献便乘上 (k)。最后对贡献求 (c) 次根，其中 (c) 是总出现次数。求贡献的最大值。
字符串长度以及贡献字符串长度之和 (le 1500)
思路
首先你需要想到我们可以全部取 (ln)，然后每次贡献就是 (+k)，求根就是 (/c)，于是问题就是最大化：
[frac{sum k}{c}
]
然后你对这个二分，判断条件就是：
[sum_{k-mid}>0
]
于是我们可以在 AC 自动机上进行dp，即设 (f_{i,j}) 表示到第 (i) 个字符串对应自动机上状态j时的最大贡献，转移显然。
于是，我们就可以在 (nllog w) 的时间复杂度内解决这个问题。
( exttt{Code})
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define MAXN 2005
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
int n,m;
int g[MAXN][MAXN][2];
double dp[MAXN][MAXN];
char s1[MAXN],s2[MAXN],ans[MAXN];
struct Auto{
double val[MAXN];
int cnt,ch[MAXN][10],fail[MAXN],sum[MAXN];
void Insert (char *s,double v){
int now = 0;
for (Int i = 1;s[i];++ i){
if (!ch[now][s[i] - '0']) ch[now][s[i] - '0'] = ++ cnt;
now = ch[now][s[i] - '0'];
}
sum[now] ++,val[now] += v;
}
void Build (){
queue <int> q;
while (!q.empty()) q.pop ();
for (Int i = 0;i < 10;++ i) if (ch[0][i]) q.push (ch[0][i]);
while (!q.empty()){
int u = q.front();q.pop ();
sum[u] += sum[fail[u]],val[u] += val[fail[u]];
for (Int i = 0;i < 10;++ i){
if (ch[u][i]) fail[ch[u][i]] = ch[fail[u]][i],q.push (ch[u][i]);
else ch[u][i] = ch[fail[u]][i];
}
}
}
double Work (double v){
for (Int i = 0;i <= cnt;++ i) val[i] -= sum[i] * v;
for (Int i = 0;i <= n;++ i) for (Int j = 0;j <= cnt;++ j) dp[i][j] = -1e6;
dp[0][0] = 0;
for (Int i = 0;i < n;++ i)
for (Int j = 0;j <= cnt;++ j)
if (dp[i][j] > -1e5)
for (Int k = 0;k < 10;++ k)
if (s1[i] == '.' || s1[i] == k + '0'){
int t = ch[j][k];
if (dp[i + 1][t] < dp[i][j] + val[t]){
dp[i + 1][t] = dp[i][j] + val[t];
g[i + 1][t][0] = k,g[i + 1][t][1] = j;
}
}
for (Int i = 0;i <= cnt;++ i) val[i] += sum[i] * v;
int pos = 0;for (Int i = 0;i <= cnt;++ i) if (dp[n][i] > dp[n][pos]) pos = i;
for (Int i = n,now = pos;i;-- i) ans[i] = g[i][now][0] + '0',now = g[i][now][1];
return dp[n][pos];
}
}T;
signed main(){
read (n,m);
scanf ("%s",s1);
for (Int i = 1,v;i <= m;++ i){
scanf ("%s",s2 + 1),read (v);
T.Insert (s2,log (v));
}
T.Build();
double l = 0,r = 1e9;
while (r - l > 1e-3){
double mid = (l + r) / 2;
if (T.Work(mid) > 0) l = mid;
else r = mid;
}
T.Work(l),printf ("%s",ans + 1);
return 0;
}