[POJ1743]Musical Theme 做题笔记

来源：互联网发布：网络教育拿学位证书编辑：程序博客网时间：2024/05/18 01:07

题目链接：http://poj.org/problem?id=1743
后缀数组+二分答案，可以看一下罗神的《后缀数组——处理字符串的有力工具》

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=20050;int n,num[N];int sa[N],rank[N],height[N];int wa[N],wb[N],wv[N],wd[N];int cmp (int *r,int a,int b,int l) {    return r[a]==r[b] && r[a+l]==r[b+l];}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++) wd[i]=0;    for (i=0;i<n;i++) wd[x[i]=r[i]]++;    for (i=1;i<m;i++) wd[i]+=wd[i-1];    for (i=n-1;i>=0;i--) sa[--wd[x[i]]]=i;    for (j=1,p=1;p<n;j<<=1,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++) wd[i]=0;        for (i=0;i<n;i++) wd[wv[i]]++;        for (i=1;i<m;i++) wd[i]+=wd[i-1];        for (i=n-1;i>=0;i--) sa[--wd[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++;    }}void calheight (int *r,int *sa,int n) {    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++);}bool valid(int len) {    int i=2,ma,mi;    while (1) {        while (i<=n&&height[i]<len)i++;        if (i>n) break;        ma=sa[i-1],mi=sa[i-1];//为什么是i-1？因为height[i]储存的是suffix[sa[i]]和suffix[sa[i-1]]的lcp        while (i<=n&&height[i]>=len) {            ma=max(ma,sa[i]);            mi=min(mi,sa[i]);            i++;        }        if (ma-mi>=len) return 1;    }    return 0;}int main () {    int ans;    while (scanf("%d",&n) && n!=0) {        for (int i=0;i<n;i++) scanf("%d",&num[i]);        if (n<10) {            puts("0");            continue;        }        n--;        for (int i=0;i<n;i++)             num[i]=num[i+1]-num[i]+89;        num[n]=0;        da(num,sa,n+1,200);        calheight(num,sa,n);        int l=1,r=n,mid;//?        while (l<r) {            mid=(l+r+1)>>1;            if (valid(mid)) l=mid;            else r=mid-1;        }        ans=l<4?0:l+1;        printf("%d\n",ans);    }    return 0;}

0 0