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  • poj3581 Sequence

    2333

    直接弃疗

    第一遍搞出sa,去最小,注意sa>=3,(从1开始),后面直接输出,然后把前面剩下的复制,加到后面就好了,为什么这样呢?? %%% http://blog.163.com/just_gogo/blog/static/1914390652011823103842787/

    还有,今天是找了一个神犇的后缀数组整理做的,感觉很爽(然后一天才4个题,感觉虚死了,代码能力什么的,,,没见过的样子)

    后缀数组题目:http://blog.csdn.net/libin56842/article/details/46440923

     1 #include <cstdio>
     2 #include <iostream>
     3 #include <cstring>
     4 #include<algorithm>
     5 #define N 100005
     6 #define inf 0x3f3f3f3f
     7 #define LL long long
     8 #define eps 1e-8
     9 using namespace std;
    10 inline int ra()
    11 {
    12     int x=0,f=1; char ch=getchar();
    13     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    14     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    15     return x*f;
    16 }
    17 
    18 const int maxn=400005;
    19 
    20 int a[maxn],num[maxn],n,len,ans[maxn];
    21 int sa[2][maxn],rank[2][maxn],v[maxn];
    22 int p=0,q=1;
    23 int find(int x)
    24 {
    25     int l=1,r=len-1;
    26     while (l<=r)
    27     {
    28         int mid=l+r>>1;
    29         if (num[mid]==x) return mid;
    30         else if (num[mid]>x) r=mid-1;
    31         else l=mid+1;
    32     }
    33 }
    34 void cal_sa(int sa[maxn], int rank[maxn], int Sa[maxn], int Rank[maxn], int k)
    35 {
    36     for (int i=1; i<=n; i++) v[rank[sa[i]]]=i;
    37     for (int i=n; i>=1; i--) if (sa[i]>k) Sa[v[rank[sa[i]-k]]--]=sa[i]-k;
    38     for (int i=n-k+1; i<=n; i++) Sa[v[rank[i]]--]=i;
    39     for (int i=1; i<=n; i++) Rank[Sa[i]]=Rank[Sa[i-1]]+(rank[Sa[i]]!=rank[Sa[i-1]] || rank[Sa[i]+k]!=rank[Sa[i-1]+k]);
    40 }
    41 void work()
    42 {
    43     for (int i=1; i<=n; i++) v[a[i]]++;
    44     for (int i=1; i<=maxn-5; i++) v[i]+=v[i-1];
    45     for (int i=1; i<=n; i++) sa[p][v[a[i]]--]=i;
    46     for (int i=1; i<=n; i++) rank[p][sa[p][i]]=rank[p][sa[p][i-1]]+(a[sa[p][i]]!=a[sa[p][i-1]]);
    47     for (int k=1; k<n; k<<=1,swap(p,q)) cal_sa(sa[p],rank[p],sa[q],rank[q],k);
    48 }
    49 void solve()
    50 {
    51     work(); int pos=0;
    52     for (int i=1; i<=n; i++) 
    53         if (sa[p][i]>=3)
    54         {
    55             pos=sa[p][i]-1;
    56             for (int j=pos+1; j<=n; j++) printf("%d
    ",ans[a[j]]);
    57             break;
    58         } // while (1);
    59     for (int i=1; i<=pos; i++) a[i+pos]=a[i];
    60     n=pos*2; memset(v,0,sizeof(v)); work();
    61 //    for (int i=1; i<=n; i++) printf("%d ",sa[p][i]); cout<<endl;
    62     for (int i=1; i<=n; i++) 
    63         if (sa[p][i]>=2 && sa[p][i]<=pos)
    64         {
    65             pos=sa[p][i]-1; //cout<<"!!"<<pos<<"!!";
    66             for (int j=pos+1; j<=n/2; j++) printf("%d
    ",ans[a[j]]);
    67             break;
    68         }
    69     for (int j=1; j<=pos; j++) printf("%d
    ",ans[a[j]]);
    70 }
    71 int main(int argc, char const *argv[])
    72 {
    73     n=ra();
    74     for (int i=n; i>=1; i--) a[i]=ra(),num[i]=a[i];
    75     sort(num+1,num+n+1); 
    76     len=unique(num+1,num+n+1)-num;
    77     for (int i=1; i<=n; i++) ans[find(a[i])]=a[i],a[i]=find(a[i]);
    78     solve();
    79     return 0;
    80 }
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  • 原文地址:https://www.cnblogs.com/ccd2333/p/6642240.html
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