题解【poj2774 Long Long Message】

Description

求两个串的最长连续公共字串

Solution

后缀数组入门题吧

把两个串连在一起，中间加一个分隔符，然后跑一遍后缀数组，得到 height 和 sa

一个 height[i] 对答案有贡献的充要条件是 sa[i] 和 sa[i-1] 分别在两个串中

Code

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 200200;
char s1[N], s2[N], S[N];
int n, tmpn, cnt[N], ans, sa[N], rk[N], height[N];  
struct node { int id, x, y; } a[N], b[N]; 
int main() {
  scanf("%s %s", s1, s2); tmpn = strlen(s1); 
  for(int i = 0; s1[i]; i++) S[++n] = s1[i]; S[++n] = '#'; 
  for(int i = 0; s2[i]; i++) S[++n] = s2[i];
  for(int i = 1; i <= n; i++) cnt[S[i]] = 1;
  for(int i = 0; i <= 128; i++) cnt[i] += cnt[i - 1];
  for(int i = 1; i <= n; i++) rk[i] = cnt[S[i]];
  for(int L = 1; L <= n; L *= 2) {
    for(int i = 1; i <= n; i++) 
      a[i].id = i, a[i].x = rk[i], a[i].y = rk[i + L];
    for(int i = 1; i <= n; i++) cnt[i] = 0;
    for(int i = 1; i <= n; i++) cnt[a[i].y]++;
    for(int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for(int i = 1; i <= n; i++) b[cnt[a[i].y]--] = a[i]; 
    for(int i = 1; i <= n; i++) cnt[i] = 0;
    for(int i = 1; i <= n; i++) cnt[a[i].x]++;
    for(int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for(int i = n; i >= 1; i--) a[cnt[b[i].x]--] = b[i]; 
    for(int i = 1; i <= n; i++) 
      if(a[i].x == a[i - 1].x && a[i].y == a[i - 1].y) 
        rk[a[i].id] = rk[a[i - 1].id];
      else rk[a[i].id] = rk[a[i - 1].id] + 1; 
  } for(int i = 1; i <= n; i++) sa[rk[i]] = i; 
  int k = 0; 
  for(int i = 1; i <= n; i++) {
    int j = sa[rk[i] - 1]; if(k) k--;
    while(i + k <= n && j + k <= n && S[i + k] == S[j + k]) k++;
    height[rk[i]] = k; 
  } for(int i = 1; i <= n; i++) 
    if(sa[i] <= tmpn && sa[i - 1] > tmpn ||
       sa[i] > tmpn && sa[i - 1] <= tmpn)
      ans = max(ans, height[i]); 
  printf("%d
", ans);
  return 0;
}