SPOJ220 relevant phrases of annihilation

http://www.spoj.com/problems/PHRASES/

Test instructions: Give n a string, find the n string inside there are 2 non-overlapping longest string length.

Idea: Two-point answer, and then you can hehe hey

PS: Spicy Chicken Topic ruined my youth, the first two points when the ANS did not assign the initial value of 0, the results will not be the answer when the output of strange digital t_t, in fact, the main blame I accidentally.

1#include <cstdio>2#include <iostream>3#include <cmath>4#include <cstring>5#include <algorithm>6 intlen,num[500005],ws[500005],wv[500005],wa[500005],wb[500005],h[500005],sa[500005],rank[500005];7 intf[500005][2],n,b[500005];8 Chars[500005];9 intRead () {Ten     intt=0, f=1;CharCh=GetChar (); One      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} A      while('0'<=ch&&ch<='9') {t=t*Ten+ch-'0'; ch=GetChar ();} -     returnt*F; - } the BOOLcmpint*r,intAintBintl) { -     returnr[a]==r[b]&&r[a+l]==r[b+l]; - } - voidDaint*r,int*sa,intNintm) { +     int*t,*x=wa,*y=wb,i,j,k,p; -      for(i=0; i<m;i++) ws[i]=0; +      for(i=0; i<n;i++) x[i]=R[i]; A      for(i=0; i<n;i++) ws[x[i]]++; at      for(i=1; i<m;i++) ws[i]+=ws[i-1]; -      for(i=n-1; i>=0; i--) sa[--ws[x[i]]]=i; -      for(j=1, p=1;p <n;j*=2, m=p) { -          for(p=0, i=n-j;i<n;i++) y[p++]=i; -          for(i=0; i<n;i++)if(SA[I]&GT;=J) y[p++]=sa[i]-J; -          for(i=0; i<m;i++) ws[i]=0; in          for(i=0; i<n;i++) wv[i]=X[y[i]]; -          for(i=0; i<n;i++) ws[wv[i]]++; to          for(i=1; i<m;i++) ws[i]+=ws[i-1]; +          for(i=n-1; i>=0; i--) sa[--ws[wv[i]]]=Y[i]; -          for(t=x,x=y,y=t,i=1, x[sa[0]]=0, p=1; i<n;i++){ theX[SA[I]]=CMP (y,sa[i-1],sa[i],j)? p1:p + +; *         } $     }Panax Notoginseng } - voidCalint*r,intN) { the     inti,j,k=0; +      for(intI=1; i<=n;i++) rank[sa[i]]=i; A      for(intI=0; i<n;h[rank[i++]]=k) { the          for(k?k--:0, j=sa[rank[i]-1];r[j+k]==r[i+k];k++); +     } - } $ BOOLCheckintmid) { $     intL,r; -      for(intI=1; i<=len;i++){ -L=i; the          while(L<=len&&h[l]<mid) l++; -R=l;Wuyi          while(R<=len&&h[r]>=mid) r++; the          for(intj=1; j<=n;j++) f[j][0]=0x3f3f3f3f, f[j][1]=-0x3f3f3f3f; -          for(intj=l-1; j<=r-1; j + +){ Wu             intk=B[sa[j]]; -             if(k==n+1)Continue; Aboutf[k][0]=std::min (f[k][0],sa[j]); $f[k][1]=std::max (f[k][1],sa[j]); -         } -         intj=0; -          for(j=1; j<=n;j++){ A             if(f[j][0]==0x3f3f3f3f) Break; +             if(f[j][1]-f[j][0]&LT;MID) Break; the         } -         if(j>n)return 1; $I=R; the     } the     return 0; the } the voidsolve () { -     intL=1, r=20000, ans=0; in      while(l<=R) { the         intMid= (l+r) >>1; the         if(Check (mid)) ans=mid,l=mid+1; About         Elser=mid-1; the     } theprintf"%d", ans); the } + intMain () { -     intt=read (); the      while(t--){Bayilen=0; the         intn=read (); the          for(intI=1; i<=n;i++){ -scanf"%s", s+1); -             intLen=strlen (s+1); the              for(intj=1; j<=len;j++) thenum[len]=s[j],b[len++]=i; the             if(i<n) num[len]=290+i,b[len++]=n+1;  the         } -num[len]=0; b[len]=n+1; theDa (num,sa,len+1,310); the cal (Num,len); then=N;94 solve (); the     } the}

SPOJ220 relevant phrases of annihilation