bzoj 4556 字符串

后缀数组，暴力硬跑
贼快

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define MAXN 100005
 7 using namespace std;
 8 int buc[MAXN],wa[MAXN],wb[MAXN];
 9 int r[MAXN],sa[MAXN],rank[MAXN],height[MAXN];
10 void getheight(int n){
11     int i,j,k=0;
12     for(i=0;i<n;i++)rank[sa[i]]=i;
13     for(i=0;i<n;height[rank[i++]]=k)
14         for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
15     return ;
16 }
17 bool cmp(int *c,int a,int b,int d)
18     {return c[a]==c[b]&&c[a+d]==c[b+d];}
19 void da(int n,int m=130){
20     int i,j,p,*x=wa,*y=wb,*t;
21     for(i=0;i<m;i++)buc[i]=0;
22     for(i=0;i<n;i++)buc[x[i]=r[i]]++;
23     for(i=1;i<m;i++)buc[i]+=buc[i-1];
24     for(i=n-1;~i;i--)sa[--buc[x[i]]]=i;
25     for(j=1,p=1;p<n;j*=2,m=p){
26         for(i=n-j,p=0;i<n;i++)y[p++]=i;
27         for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
28         for(i=0;i<m;i++)buc[i]=0;
29         for(i=0;i<n;i++)buc[x[y[i]]]++;
30         for(i=1;i<m;i++)buc[i]+=buc[i-1];
31         for(i=n-1;~i;i--)sa[--buc[x[y[i]]]]=y[i];
32         for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
33             x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
34     }
35     getheight(n);
36     return ;
37 }
38 int n,q;
39 char s[MAXN];
40 int main(){
41     scanf("%d%d",&n,&q);
42     scanf("%s",s);
43     for(int i=0;i<n;i++)r[i]=s[i];
44     da(n+1);
45     int ans,minn,pos,now,a,b,c,d;
46     while(q--){
47         scanf("%d%d%d%d",&a,&b,&c,&d);
48         a--;b--;c--;d--;
49         ans=0; minn=min(d-c+1,b-a+1); 
50         pos=rank[c]; 
51         if(sa[pos]>=a&&sa[pos]<=b)ans=max(ans,min(minn,min(b-sa[pos]+1,d-c+1)));
52         minn=min(minn,height[pos]);
53         for(int i=pos-1;i>=0;i--){
54             if(minn<=ans)break;
55             now=sa[i];
56             if(now>=a&&now<=b)ans=max(ans,min(minn,min(b-sa[i]+1,d-c+1)));
57             minn=min(minn,height[i]);
58         }
59         minn=min(d-c+1,b-a+1);
60         for(int i=pos+1;i<=n;i++){
61             minn=min(minn,height[i]);
62             if(minn<=ans)break;
63             now=sa[i];
64             if(now>=a&&now<=b){ans=max(ans,min(minn,min(b-sa[i]+1,d-c+1)));}
65         }
66         printf("%d\n",ans);
67     }
68     return 0;
69 }
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posted @ 2017-09-26 17:48  Ren_Ivan  阅读(107)  评论(0)  编辑  收藏  举报