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hdu 3068 最长回文子串 TLE

后缀数组+RMQ是O(nlogn)的,会TLE.....

标准解法好像是马拉车,O(n)....

  1 #include "algorithm"
  2 #include "cstdio"
  3 #include "cstring"
  4 using namespace std;
  5 #define maxn 220020
  6 
  7 int wa[maxn],wb[maxn],wv[maxn],ws[maxn];
  8 int rank[maxn],height[maxn];
  9 int r[maxn],sa[maxn],RMQ[maxn][20];
 10 char s[110010];
 11 int n;
 12 
 13 int cmp(int *r,int a,int b,int l)
 14 {
 15     return r[a]==r[b]&&r[a+l]==r[b+l];
 16 }
 17 
 18 void da(int *r,int *sa,int n,int m)
 19 {
 20     int i,j,p,*x=wa,*y=wb,*t;
 21     for(i=0; i<m; i++) ws[i]=0;     //注意此处有个小问题:ws和iostream里面某地重名了..
 22     for(i=0; i<n; i++) ws[x[i]=r[i]]++;
 23     for(i=1; i<m; i++) ws[i]+=ws[i-1];
 24     for(i=n-1; i>=0; i--) sa[--ws[x[i]]]=i;
 25     for(j=1,p=1; p<n; j*=2,m=p)
 26     {
 27 
 28         for(p=0,i=n-j; i<n; i++) y[p++]=i;
 29         for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
 30         for(i=0; i<n; i++) wv[i]=x[y[i]];
 31         for(i=0; i<m; i++) ws[i]=0;
 32         for(i=0; i<n; i++) ws[wv[i]]++;
 33         for(i=1; i<m; i++) ws[i]+=ws[i-1];
 34         for(i=n-1; i>=0; i--) sa[--ws[wv[i]]]=y[i];
 35         for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
 36             x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
 37     }
 38     return;
 39 }
 40 
 41 void calheight(int *r,int *sa,int n)
 42 {
 43     int i,j,k=0;
 44     for(i=1; i<=n; i++) rank[sa[i]]=i;
 45     for(i=0; i<n; height[rank[i++]]=k)
 46         for(k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);
 47     return;
 48 }
 49 
 50 bool _same(int lx,int ly,int l1,int l2)
 51 {
 52     return (((lx<=l1-1) && (ly>=l1+1))||((ly<=l1-1) && (lx>=l1+1)));
 53 }
 54 
 55 void ST()        //初始化
 56 {
 57     memset(RMQ,1,sizeof(RMQ));
 58     for(int i=1;i<=n;i++)
 59         RMQ[i][0]=height[i];
 60     for(int j=1;(1<<j)<=n;j++)
 61         for(int i=1;i+(1<<j)-1<=n;i++)
 62         RMQ[i][j]=min(RMQ[i][j-1],RMQ[i+(1<<(j-1))][j-1]);
 63 }
 64 
 65 int Query(int L,int R)    //求a[L..R]区间的最值
 66 {
 67     int k=0;
 68     while((1<<(k+1))<=R-L+1) k++;
 69     int tb=min(RMQ[L][k],RMQ[R-(1<<k)+1][k]);
 70     return tb;
 71 }
 72 
 73 int calc(int l,int r)       //求l开始后缀和r开始后缀的最长公共前缀
 74 {
 75     int tl=rank[l],tr=rank[r];
 76     if (tl>tr)      swap(tl,tr);
 77     tl++;
 78     int ans=Query(tl,tr);       //相当于RMQ问题
 79 //  printf("calc:  %d %d -- %d %d == %d\n",l,r,tl,tr,ans);
 80     return ans;
 81 }
 82 
 83 int _max(int a,int b,int c,int d)
 84 {
 85     int mx=a;
 86     if (b>mx) mx=b;
 87     if (c>mx) mx=c;
 88     if (d>mx) mx=d;
 89     return mx;
 90 }
 91 
 92 //da(r,sa,n+1,128);
 93 //calheight(r,sa,n);
 94 int main()
 95 {
 96     while(~scanf("%s",s))
 97     {
 98         n=strlen(s);
 99 
100         for (int i=0; i<n; i++)
101             r[i]=int(s[i])-int('a')+1;
102 
103         r[n]=100;
104         for (int i=n-1; i>=0; i--)
105             r[2*n-i]=int(s[i])-int('a')+1;
106         r[2*n+1]=0;
107         int n2=n;       n=2*n+1;
108 
109 //      for (int i=0; i<=n; i++)      printf("%d ",r[i]);
110 //        printf("\n %d   %d\n",n,n2);
111 
112         da(r,sa,n+1,128);
113         calheight(r,sa,n);
114 
115 //      for (int i=0; i<=n; i++)
116 //          printf("%d  %d  %d\n",sa[i],height[i],rank[i]);
117 //      printf("\n");
118 
119         ST();
120         int ans=1;
121         for (int i=1; i<=n2-2; i++)
122         {
123             int t1=calc(i,n-i)*2;
124             int t2=calc(i+1,n-i)*2+1;
125             int t3=calc(i+1,n-i-1)*2;
126 //          printf("%d %d %d\n",t1,t2,t3);
127             ans=_max(ans,t1,t2,t3);
128         }
129         printf("%d\n",ans);
130     }
131     return 0;
132 }

 

 

 

 

 

 

 

 

 

最近两天有点不在状态....先滚去整理模板吧

posted on 2014-10-25 18:04  Pentium.Labs  阅读(332)  评论(0编辑  收藏  举报



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