poj 3261 Milk Patterns 后缀数组

题目链接

求可重叠k次的最长子串。

二分子串长度x, 将height数组分组, 看是否有一组中的数量大于等于k。

  1 #include <iostream>
  2 #include <vector>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <map>
  8 #include <set>
  9 #include <string>
 10 #include <queue>
 11 #include <stack>
 12 #include <bitset>
 13 using namespace std;
 14 #define pb(x) push_back(x)
 15 #define ll long long
 16 #define mk(x, y) make_pair(x, y)
 17 #define lson l, m, rt<<1
 18 #define mem(a) memset(a, 0, sizeof(a))
 19 #define rson m+1, r, rt<<1|1
 20 #define mem1(a) memset(a, -1, sizeof(a))
 21 #define mem2(a) memset(a, 0x3f, sizeof(a))
 22 #define rep(i, n, a) for(int i = a; i<n; i++)
 23 #define fi first
 24 #define se second
 25 typedef pair<int, int> pll;
 26 const double PI = acos(-1.0);
 27 const double eps = 1e-8;
 28 const int mod = 1e9+7;
 29 const int inf = 1061109567;
 30 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 31 const int maxn = 2e4+5;
 32 int sa[maxn];
 33 int t1[maxn],t2[maxn],c[maxn], a[maxn];
 34 int rankk[maxn],height[maxn];
 35 void build_sa(int s[],int n,int m)
 36 {
 37     int i,j,p,*x=t1,*y=t2;
 38     for(i=0;i<m;i++)c[i]=0;
 39     for(i=0;i<n;i++)c[x[i]=s[i]]++;
 40     for(i=1;i<m;i++)c[i]+=c[i-1];
 41     for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
 42     for(j=1;j<=n;j<<=1)
 43     {
 44         p=0;
 45         for(i=n-j;i<n;i++)y[p++]=i;
 46         for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
 47         for(i=0;i<m;i++)c[i]=0;
 48         for(i=0;i<n;i++)c[x[y[i]]]++;
 49         for(i=1;i<m;i++)c[i]+=c[i-1];
 50         for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
 51         swap(x,y);
 52         p=1;x[sa[0]]=0;
 53         for(i=1;i<n;i++)
 54             x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
 55         if(p>=n)break;
 56         m=p;
 57     }
 58 }
 59 void getHeight(int s[],int n)
 60 {
 61     int i,j,k=0;
 62     for(i=0;i<=n;i++)rankk[sa[i]]=i;
 63     for(i=0;i<n;i++)
 64     {
 65         if(k)k--;
 66         j=sa[rankk[i]-1];
 67         while(s[i+k]==s[j+k])k++;
 68         height[rankk[i]]=k;
 69     }
 70 }
 71 int check(int mid, int n, int m) {
 72     int cnt = 1;
 73     for(int i = 1; i<=n; i++) {
 74         if(height[i]>=mid) {
 75             cnt++;
 76             if(cnt>=m)
 77                 return 1;
 78         } else {
 79             cnt = 1;
 80         }
 81     }
 82     return 0;
 83 }
 84 int main()
 85 {
 86     int n, m;
 87     while(cin>>n>>m) {
 88         int maxx = 0;
 89         for(int i = 0; i<n; i++) {
 90             scanf("%d", &a[i]);
 91             maxx = max(maxx, a[i]);
 92         }
 93         a[n] = 0;
 94         build_sa(a, n+1, maxx+1);
 95         getHeight(a, n);
 96         int l = 1, r = n, ans;
 97         while(l<=r) {
 98             int mid = l+r>>1;
 99             if(!check(mid, n, m)) {
100                 r = mid-1;
101             } else {
102                 l = mid+1;
103                 ans = mid;
104             }
105         }
106         cout<<ans<<endl;
107     }
108     return 0;
109 }
原文地址:https://www.cnblogs.com/yohaha/p/5098131.html