HDU 6153 A Secret (KMP)

Posted 2020-10-03 dwtfukgv

题意：给定两个串，求其中一个串 s 的每个后缀在另一个串 t 中出现的次数。

析：首先先把两个串进行反转，这样后缀就成了前缀。然后求出 s 的失配函数，然后在 t 上跑一遍，如果发现不匹配了或者是已经完全匹配了，要计算，前面出现了多少个串的长度匹配也就是 1 + 2 + 3 + .... + j 在 j 处失配，然后再进行失配处理。注意别忘了，匹配完还要再加上最后的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 100;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
    return r > 0 && r <= n && c > 0 && c <= m;
}

char s[maxn], t[maxn];
int f[maxn];

void getFail(){
  f[0] = f[1] = 0;
  for(int i = 1; i < n; ++i){
    int j = f[i];
    while(j && s[j] != s[i])  j = f[j];
    f[i+1] = s[i] == s[j] ? j+1 : 0;
  }
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%s", t);
    int len = strlen(t);
    reverse(t, t + len);
    scanf("%s", s);
    n = strlen(s);
    reverse(s, s + n);
    getFail();
    LL ans = 0;
    int j = 0;
    for(int i = 0; i < len; ++i){
      while(j && s[j] != t[i]){
        ans = (ans + (LL)j * (j+1LL) / 2) % mod;
        j = f[j];
      }
      if(s[j] == t[i])  ++j;
      if(j == n){
        ans = (ans + (LL)j * (j+1LL) / 2 ) % mod;
        j = f[j];
      }
    }
    while(j){
      ans = (ans + (LL)j * (j+1LL) / 2) % mod;
      j = f[j];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}