HDU 4117 GRE words (suffix array + dp)

Question:

The maximum number of words that can be remembered is the sum.

Remember that the previous word is a substring of the word.

Train of Thought Analysis:

First, you must declare that the question is data water before using SA.

The complexity of SA can reach orz (sumlen * sumlen) at most )...

Therefore, the SA processes whether the string is the sub-string of the Next string, and if so, the transfer equation.

DP [I] = max (DP [I], DP [J] + val [I])...

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#define maxn 350005#define inf 0x3f3f3f3fusing namespace std;int str[maxn];int sa[maxn],t1[maxn],t2[maxn],c[maxn],n;int bel[maxn];int st[maxn],sumlen[maxn],val[maxn];int dp[maxn];void suffix(int m){    int *x=t1,*y=t2;    for(int i=0;i<m;i++)c[i]=0;    for(int i=0;i<n;i++)c[x[i]=str[i]]++;    for(int i=1;i<m;i++)c[i]+=c[i-1];    for(int i=n-1;i>=0;i--)sa[--c[x[i]]]=i;    for(int k=1;k<=n;k<<=1)    {        int p=0;        for(int i=n-k;i<n;i++)y[p++]=i;        for(int i=0;i<n;i++)if(sa[i]>=k)y[p++]=sa[i]-k;        for(int i=0;i<m;i++)c[i]=0;        for(int i=0;i<n;i++)c[x[y[i]]]++;        for(int i=0;i<m;i++)c[i]+=c[i-1];        for(int i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;x[sa[0]]=0;        for(int i=1;i<n;i++)        x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;        if(p>=n)break;        m=p;    }}int rank[maxn],height[maxn];void getheight(){    int k=0;    for(int i=0;i<n;i++)rank[sa[i]]=i;    for(int i=0;i<n;i++)    {        if(k)k--;        if(!rank[i])continue;        int j=sa[rank[i]-1];        while(str[i+k]==str[j+k])k++;        height[rank[i]]=k;    }}char tmp[maxn];int main(){    int T,cas=1;    for(scanf("%d",&T);T--;)    {        int N;        scanf("%d",&N);        n=0;        for(int i=1;i<=N;i++)        {            scanf("%s%d",tmp,&val[i]);            dp[i]=val[i];            int len=strlen(tmp);            st[i]=n;            sumlen[i]=len;            for(int j=0;j<len;j++)            {                str[n]=tmp[j];                bel[n++]=i;            }            bel[n]=i;            str[n++]=i+128;        }        str[n-1]=0;        suffix(30000);        getheight();        for(int i=1;i<=N;i++)        {            int mlen=inf;            for(int j=rank[st[i]]+1;j<n;j++)            {                mlen=min(mlen,height[j]);                if(mlen<sumlen[i])break;                if(bel[sa[j]]>i)                {                    dp[bel[sa[j]]]=max(dp[bel[sa[j]]],val[bel[sa[j]]]+dp[i]);                }            }            mlen=inf;            for(int j=rank[st[i]]-1;j>0;j--)            {                mlen=min(mlen,height[j+1]);                if(mlen<sumlen[i])break;                if(bel[sa[j]]>i)                {                    dp[bel[sa[j]]]=max(dp[bel[sa[j]]],val[bel[sa[j]]]+dp[i]);                }            }        }        printf("Case #%d: %d\n",cas++,max(0,*max_element(dp+1,dp+1+N)));    }    return 0;}/*16abbab 800a 1ab 2abb 3baba 5abbab 8*/

HDU 4117 GRE words (suffix array + dp)