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  • URAL 1297 Palindrome (后缀数组+RMQ)

    题意:给定一个字符串,求一个最长的回回文子串,多解输出第一个。

    析:把字符串翻转然后放到后面去,中间用另一个字符隔开,然后枚举每一个回文串的的位置,对第 i 个位置,那么对应着第二个串的最长公共前缀,

    求最长公共子串,可以用RMQ解决。

    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>
    #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 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 = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 2000 + 10;
    const int mod = 1000007;
    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;
    }
    
    struct Array{
      int s[maxn], sa[maxn], t[maxn], t2[maxn];
      int h[maxn], r[maxn], c[maxn];
      int n;
      int dp[maxn][20];
    
      void init(){ n = 0;  memset(sa, 0, sizeof sa); }
      void build_sa(int m){
        int *x = t, *y = t2;
        for(int i = 0; i < m; ++i)  c[i] = 0;
        for(int i = 0; i < n; ++i)  ++c[x[i] = s[i]];
        for(int i = 1; i < m; ++i)  c[i] += c[i-1];
        for(int i = n-1; i >= 0; --i)  sa[--c[x[i]]] = i;
    
        for(int k = 1; k <= n; k <<= 1){
          int p = 0;
          for(int i = n-k; i < n; ++i)  y[p++] = i;
          for(int i = 0; i < n; ++i)  if(sa[i] >= k)  y[p++] = sa[i] - k;
          for(int i = 0; i < m; ++i)  c[i] = 0;
          for(int i = 0; i < n; ++i)  ++c[x[y[i]]];
          for(int i = 1; i < m; ++i)  c[i] += c[i-1];
          for(int i = n-1; i >= 0; --i)  sa[--c[x[y[i]]]] = y[i];
    
          swap(x, y);
          p = 1;  x[sa[0]] = 0;
          for(int i = 1; i < n; ++i)
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
          if(p >= n)  break;
          m = p;
        }
      }
    
      void getHight(){
        int k = 0;
        for(int i = 0; i < n; ++i)  r[sa[i]] = i;
        for(int i = 0; i < n; ++i){
          if(k)  --k;
          int j = sa[r[i]-1];
          while(s[i+k] == s[j+k])  ++k;
          h[r[i]] = k;
        }
      }
    
      void rmq_init(){
        for(int i = 1; i <= n; ++i)  dp[i][0] = h[i];
        for(int j = 1; (1<<j) <= n; ++j)
          for(int i = 1; i + (1<<j) <= n; ++i)
            dp[i][j] = min(dp[i][j-1], dp[i+(1<<j-1)][j-1]);
      }
    
      int query(int L, int R){
        L = r[L], R = r[R];
        if(L > R)  swap(L, R);
        ++L;
        int k = int(log(R-L+1) / log(2.0));
        return min(dp[L][k], dp[R-(1<<k)+1][k]);
      }
    };
    char s[maxn];
    Array arr;
    
    int main(){
      while(scanf("%s", s) == 1){
        n = strlen(s);
        arr.init();
        for(int i = 0; i < n; ++i)  arr.s[arr.n++] = s[i];
        arr.s[arr.n++] = 128;
        for(int i = n-1; i >= 0; --i) arr.s[arr.n++] = s[i];
        arr.s[arr.n++] = 0;
        arr.build_sa(130);
        arr.getHight();
        arr.rmq_init();
        int ans = 0, idx;
        for(int i = 0; i < n; ++i){
          int res = max(arr.query(i, 2*n-i) * 2 - 1, arr.query(i, 2*n-i+1) * 2); // odd
          if(res > ans){
            ans = res;  idx = i;
          }
        }
        if(ans & 1)  for(int i = idx-ans/2; i <= ans/2+idx; ++i)
          printf("%c", s[i]);
        else for(int i = idx-ans/2; i < idx+ans/2; ++i)
          printf("%c", s[i]);
        printf("
    ");
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6705640.html
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