CF 494B - Obsessive String

rig[i]表示以i开头的满足条件的串(必须含有i)

转移方程

for (int j=p+1;j<n-1;j++){
  for (int k=j+1;k<n;k++){
       rig[i] += rig[k];
  }
  rig[i]++;
}
rig[i]++;

t是s[i..p]的字串且|p-i|最小

可以用KMP预处理出来所有可以匹配的后缀点,然后二分出p的位置

接下来就是将上式优化

令 sum[i] = sgma(rig[j]) (j>=i&&j<n)

ans[i] = sgma(sum[j]) (j>=i&&j<n)

所以方程就是

rig[i] = (ans[j+1] + n-j);
sum[i] = (sum[i+1] + rig[i]);
ans[i] = (ans[i+1] + sum[i]);

代码如下

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <vector>
 4 using namespace std;
 5 #define N 100000
 6 vector<int> gp;
 7 char s[N+20],t[N+20];
 8 int p[N+20];
 9 long long rig[N+20],sum[N+20],ans[N+20];
10 const int mod = (int)(1e9+7);
11 void get_next(char *str)
12 {
13     int i = 0,j = -1, l = strlen(str);
14     p[0] = -1;
15     while (i<l)
16     {
17         if (j==-1||str[i]==str[j]) p[++i] = ++j;
18         else j = p[j];
19     }
20 }
21 bool KMP(char *s1,char *s2)  //串s1中寻找串s2
22 {
23     int l1 = strlen(s1), l2 = strlen(s2), i = 0, j = 0;
24     while (i<l1&&j<l2)
25     {
26         if (j==-1||s1[i]==s2[j]) i++,j++;
27         else j= p[j];
28         if (j==l2){
29             gp.push_back(i-1);
30             j = p[j];
31         }
32     }
33 }
34 int main(){
35     scanf("%s%s",s,t);
36     get_next(t);
37     KMP(s,t);
38     int n = strlen(s);
39     int m = strlen(t);
40     for (int i=n-1;i>=0;i--){
41         vector<int>::iterator it = lower_bound(gp.begin(),gp.end(),i+m-1);
42         if (it!=gp.end()){
43             int j = *it;
44             rig[i] = (ans[j+1] + n-j) % mod;
45             sum[i] = (sum[i+1] + rig[i]) % mod;
46             ans[i] = (ans[i+1] + sum[i]) % mod;
47         }
48     }
49     long long res = 0;
50     for (int i=0;i<n;i++) res = (res + rig[i]) % mod;
51     printf("%I64d\n",res);
52     return 0;
53 }