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这道是求长度不小于 k 的公共子串的个数...很不幸,我又TLE了...
解法参考论文以及下面的链接
http://www.cnblogs.com/vongang/archive/2012/11/20/2778481.html
http://hi.baidu.com/fpkelejggfbfimd/item/5c76cfcba28fba26e90f2ea6
1 const maxn=200419; 2 var 3 c,h,rank,sa,x,y,stack:array[0..maxn] of longint; 4 n,k,top:longint; 5 a,b,d:int64; 6 s1,s2:ansistring; 7 8 function max(x,y:longint):longint; begin if x>y then exit(x) else exit(y); end; 9 function min(x,y:longint):longint; begin if x<y then exit(x) else exit(y); end; 10 11 procedure make; 12 var p,i,tot:longint; 13 begin 14 p:=1; 15 while p<n do 16 begin 17 fillchar(c,sizeof(c),0); 18 for i:= 1 to n-p do y[i]:=rank[i+p]; 19 for i:= n-p+1 to n do y[i]:=0; 20 for i:= 1 to n do inc(c[y[i]]); 21 for i:= 1 to n do inc(c[i],c[i-1]); 22 for i:= 1 to n do 23 begin 24 sa[c[y[i]]]:=i; 25 dec(c[y[i]]); 26 end; 27 fillchar(c,sizeof(c),0); 28 for i:= 1 to n do x[i]:=rank[i]; 29 for i:= 1 to n do inc(c[x[i]]); 30 for i:= 1 to n do inc(c[i],c[i-1]); 31 for i:= n downto 1 do 32 begin 33 y[sa[i]]:=c[x[sa[i]]]; 34 dec(c[x[sa[i]]]); 35 end; 36 for i:= 1 to n do sa[y[i]]:=i; 37 tot:=1; 38 rank[sa[1]]:=1; 39 for i:= 2 to n do 40 begin 41 if (x[sa[i]]<>x[sa[i-1]]) or (x[sa[i]+p]<>x[sa[i-1]+p]) then inc(tot); 42 rank[sa[i]]:=tot; 43 end; 44 p:=p<<1; 45 end; 46 for i:= 1 to n do sa[rank[i]]:=i; 47 end; 48 49 procedure makeh(s:ansistring); 50 var i,j,p:longint; 51 begin 52 h[1]:=0; p:=0; 53 for i:= 1 to n do 54 begin 55 p:=max(p-1,0); 56 if rank[i]=1 then continue; 57 j:=sa[rank[i]-1]; 58 while (i+p<=n) and (j+p<=n) and (s[i+p]=s[j+p]) do inc(p); 59 h[rank[i]]:=p; 60 end; 61 end; 62 63 procedure init(s:ansistring); 64 var i,tot:longint; 65 ch:char; 66 begin 67 n:=length(s); 68 for i:= 1 to n do c[i]:=0; 69 for i:= 1 to n do x[i]:=ord(s[i]); 70 for i:= 1 to n do inc(c[x[i]]); 71 for i:= 1 to 122 do inc(c[i],c[i-1]); 72 for i:= 1 to n do 73 begin 74 sa[c[x[i]]]:=i; 75 dec(c[x[i]]); 76 end; 77 rank[sa[1]]:=1; 78 tot:=1; 79 for i:= 2 to n do 80 begin 81 if x[sa[i]]<>x[sa[i-1]] then inc(tot); 82 rank[sa[i]]:=tot; 83 end; 84 fillchar(x,sizeof(x),0); 85 fillchar(y,sizeof(y),0); 86 make; 87 makeh(s); 88 end; 89 90 function calc(s:ansistring):int64; 91 var ctb,i,m,ht,top:longint; 92 ans:int64; 93 begin 94 ans:=0; 95 init(s); 96 h[0]:=k-1; h[n+1]:=k-1; 97 top:=0; i:=1; 98 stack[0]:=0; 99 while i<=n+1 do 100 begin 101 ht:=h[stack[top]]; 102 if ((h[i]<k) and (top=0)) or (h[i]=ht) then inc(i) 103 else if h[i]>ht then begin inc(top);stack[top]:=i; inc(i); end 104 else 105 begin 106 m:=i-stack[top]+1; 107 if (h[i]>=k) and (h[i]>h[stack[top-1]]) then 108 begin 109 ctb:=ht-h[i]; 110 h[stack[top]]:=h[i]; 111 end 112 else 113 begin 114 ctb:=ht-h[stack[top-1]]; 115 dec(top); 116 end; 117 inc(ans,(int64(m)*int64(m-1) div 2) *int64(ctb)); 118 end 119 end; 120 exit(ans); 121 end; 122 123 Begin 124 readln(k); 125 while k<>0 do 126 begin 127 readln(s1); 128 a:=calc(s1); 129 readln(s2); 130 b:=calc(s2); 131 s1:=s1+‘$‘+s2; 132 d:=calc(s1); 133 writeln(d-a-b); 134 readln(k); 135 end; 136 End.
【POJ3415】 Common Substrings (SA+单调栈)
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原文地址:http://www.cnblogs.com/EC-Ecstasy/p/4174671.html