Hdu 4080 & Poj 3882 Stammering Aliens (后缀数组可重叠k次最长重复子串)

╰半夏微凉° 2021-11-11 15:20 252阅读 0赞

和Poj3261差不多：[POJ 3261 Milk Patterns (离散化+后缀数组 可重叠k次最长重复子串) - whyorwhnt的专栏][POJ 3261 Milk Patterns _ _k_ - whyorwhnt]

利用二分处理。需要输出最靠右的起始位置。与上题不同的是，上题只需确定长度，所以二分判定时满足条件直接返回；本题需要在二分的时候计算出起始位置，所以必须完成对整个height数组的扫描后才能返回。

对第一份代码中判定部分的说明：因为height\[i\]对应sa\[i-1\]和sa\[i\]，所以要从i-1处开始处理。

附一组测试数据：

In  
2  
dcdcbaba  
  
Out  
2 6

感觉本题测试数据不强，一些有小漏洞的代码也能AC……

#include <cstdio>
    #include <cstring>
    #define max(a,b) ((a)>(b)?(a):(b))
    #define min(a,b) ((a)<(b)?(a):(b))
    
    const int N = int(2e5)+10;
    
    int cmp(int *r,int a,int b,int l){
        return (r[a]==r[b]) && (r[a+l]==r[b+l]);
    }
    int wa[N],wb[N],ws[N],wv[N];
    int rank[N],height[N];
    
    void DA(int *r,int *sa,int n,int m){
        int i,j,p,*x=wa,*y=wb,*t;
        for(i=0;i<m;i++) ws[i]=0;
        for(i=0;i<n;i++) ws[x[i]=r[i]]++;
        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]>=j) y[p++]=sa[i]-j;
            for(i=0;i<n;i++) wv[i]=x[y[i]];
            for(i=0;i<m;i++) ws[i]=0;
            for(i=0;i<n;i++) ws[wv[i]]++;
            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,p=1,x[sa[0]]=0,i=1;i<n;i++)
                x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
            //printf("p = %d\n", p );
        }
    }
    void calheight(int *r,int *sa,int n){
      //  memset(height,0,sizeof(height));
      //  memset(rank,0,sizeof(rank));
        int i,j,k=0;
        for(i=1;i<=n;i++) rank[sa[i]]=i;
        for(i=0;i<n; height[rank[i++]] = k )
        for(k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);
    }
    
    
    int data[N],sa[N],temp[N],n,k,start;
    char str[N];
    
    bool Judge (int x)  
    {  
    	bool flag=false;
    	start=-1;
        for (int i=1;i<=n;i++)  
        {  
            int L=i;  
            while (L<=n && height[L]<x) L++;  
            if (L>n) break;  
            int R=L;  
            while (R<=n && height[R]>=x) R++;  
    		if (R-L+1>=k)
            {
                for (int j=L-1;j<=R-1;j++)
    				start=max(start,sa[j]);
                flag=true;
            }
            i=R;  
        }  
        return flag;  
    } 
    
    void Deal ()
    {
        DA(data,sa,n+1,200);
        calheight(data,sa,n);
    	int low=0,high=n,mid,ans=0,s;
    	while (low<high)   //注意二分的写法
    	{
    		mid = (low+high)>>1;
    		if ( Judge(mid) )
    		{
    			ans=mid;
    			low=mid+1;
    			s=start;
    		}
    		else high = mid;
    	}
    	if (ans==0)
    		printf("none\n");
    	else
    		printf("%d %d\n",ans,s);
    }
    
    int main ()
    {  
    #ifdef ONLINE_JUDGE
    #else
    	freopen("read.txt","r",stdin);
    #endif
    	while (~scanf("%d",&k),k)
    	{
    		scanf("%s",str);
    		n=strlen(str);
    		if (k==1)
    		{
    			printf("%d %d\n",n,0);  
    			continue;  
    		}		
    		for (int i=0;i<n;i++)
    			data[i]=(int)str[i];
    		data[n] = 0;
            Deal();
    	}
    	return 0;
    }

#include <cstdio>
    #include <cstring>
    #define max(a,b) ((a)>(b)?(a):(b))
    #define min(a,b) ((a)<(b)?(a):(b))
    
    const int N = int(4e4)+10;
    
    #define F(x) ((x)/3+((x)%3==1?0:tb))
    #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
    
    int wa[N],wb[N],wv[N],ws[N];
    int c0 (int *r,int a,int b){
    	return r[a]==r[b] && r[a+1]==r[b+1] && r[a+2]==r[b+2];
    }
    int c12 (int k,int *r,int a,int b){
    	if (k==2) return r[a]<r[b] || r[a]==r[b] && c12(1,r,a+1,b+1);
    	else return r[a]<r[b] || r[a]==r[b] && wv[a+1]<wv[b+1];
    }
    void sort (int *r,int *a,int *b,int n,int m){
    	int i;
    	for(i=0;i<n;i++) wv[i]=r[a[i]];
    	for(i=0;i<m;i++) ws[i]=0;
    	for(i=0;i<n;i++) ws[wv[i]]++;
    	for(i=1;i<m;i++) ws[i]+=ws[i-1];
    	for(i=n-1;i>=0;i--) b[--ws[wv[i]]]=a[i];
    }
    void DC3 (int *r,int *sa,int n,int m){
    	int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;
    	r[n]=r[n+1]=0;
    	for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;
    	sort(r+2,wa,wb,tbc,m);
    	sort(r+1,wb,wa,tbc,m);
    	sort(r,wa,wb,tbc,m);
    	for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)
    		rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;
    	if(p<tbc) DC3(rn,san,tbc,p);
    	else for(i=0;i<tbc;i++) san[rn[i]]=i;
    	for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;
    	if(n%3==1) wb[ta++]=n-1;
    	sort(r,wb,wa,ta,m);
    	for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;
    	for(i=0,j=0,p=0;i<ta && j<tbc;p++)
    		sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];
    	for(;i<ta;p++) sa[p]=wa[i++];
    	for(;j<tbc;p++) sa[p]=wb[j++];
    }  
    int rank[N],height[N],sa[3*N],data[3*N];
    
    void calheight(int *r,int *sa,int n){
    //	memset(height,0,sizeof(height));
    //	memset(rank,0,sizeof(rank));
    	int i,j,k=0;
    	for(i=1;i<=n;i++) rank[sa[i]]=i;
    	for(i=0;i<n; height[rank[i++]] = k )
    	for(k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);
    }
    
    int n,k,start;
    char str[N];
    
    bool Judge (int mid)
    {
    	int cnt=1;
    	start=-1;
    	int tmp=sa[1];   //存储每一个分组中,sa的最大值
    	bool flag=false;
    	for (int i=2;i<=n;i++)
    		if (height[i]<mid)
    		{
    			tmp=sa[i];
    			cnt=1;
    		}
    		else
    		{
    			cnt++;
    			tmp=max(tmp,sa[i]);
    			if (cnt >= k)
    			{
    				start=max(start,tmp);
    				flag=true;
    			}
    		}
    	return flag;
    }
    
    void Deal ()  
    {  
    	DC3(data,sa,n+1,128);  
    	calheight(data,sa,n);  
    	int low=0,high=n,mid,ans=0,s;  
    	while (low<high)   //注意二分的写法  
    	{  
    		mid = (low+high)>>1;  
    		if ( Judge(mid) )  
    		{
    			ans=mid;  
    			low=mid+1;  
    			s=start;  
    		}  
    		else high = mid;  
    	}  
    	if (ans==0)  
    		printf("none\n");  
    	else  
    		printf("%d %d\n",ans,s);  
    }
    
    int main ()
    {
    	while (scanf("%d%d",&k),k)
    	{
    		scanf("%s",str);
    		n=strlen(str);
    		if (k==1)  
    		{
    			printf("%d %d\n",n,0);    
    			continue;    
    		}
    		for (int i=0;i<n;i++)
    			data[i]=(int)str[i];
    		data[n] = 0;
    		Deal();
    	}
    	return 0;
    }
    /*
    In
    2
    dcdcbaba
    
    Out
    2 6
    */

[POJ 3261 Milk Patterns _ _k_ - whyorwhnt]: http://blog.csdn.net/whyorwhnt/article/details/9771673