POJ1743Musical Theme求解不重叠的最长子串长度（后缀数组+二分求解）

题目链接http://poj.org/problem?id=1743

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题目中介绍相差a个数字形成一个theme，一开始没看清楚题以为是只要子串正反匹配就可以，只是将字符串连接起来求解最长子串长度，自习读题后才发现错误，然后发现只需求出两相邻数字差值，然后求解最长子串长度，然后答案+1便可，求解不重叠的最长子串长度，利用height数组求解出后缀数组的最长公共前缀长度，然后二分枚举去判读答案（在满足答案时求解两者距离最大值即sa[i]数组满足最大值与最小值的差值）

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/********************    Acm ID: OffensiveChild    Author: wzh    proverbs£ºyiroad keep straight on    Tittle:Musical Theme    Time:2016.07.29    solutions:*********************/#include <iostream>#include<algorithm>#include<cstdio>#include<string>#include<cstring>#include<cmath>#include<vector>#include<set>#include<map>#include<stack>using namespace std;typedef long long ll;const int maxm=1000+5;const int maxn=60000+5;const int INF=0x3f3f3f3f;const double eps=1e-8;int num[maxn];int sa[maxn], Rank[maxn], height[maxn];int wa[maxn], wb[maxn], wv[maxn], wd[maxn];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 n, int m)           //  倍增算法 r为待匹配数组  n为总长度 m为字符范围{    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 *= 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 ++) 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 n)            //  求height数组。{    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 check(int n,int k){    int amax=sa[1],amin=sa[1];    for(int i=2; i<=n;i++)    {        if(height[i]<k)amax=amin=sa[i];//当满足height[i]==k是求最大长度的解        else        {            amax=max(amax,sa[i]);            amin=min(amin,sa[i]);            if(amax-amin>k)return true;        }    }    return false;}int anum[maxn];int main(){    int n;    while(~scanf("%d",&n))    {        if(n==0)break;        memset(num,0,sizeof(num));        memset(anum,0,sizeof(anum));        for(int i=0; i<n; i++)        {            scanf("%d",&num[i]);        }        for(int i=n-1; i>0; i--)        {            anum[i]=num[i]-num[i-1]+90;        }        n--;        for(int i=0;i<n;i++)        {            anum[i]=anum[i+1];        }        anum[n]=0;        da(anum,n+1,200);        calHeight(anum,n);        int ans=0;        int l=1,r=n/2;        while(l<=r)        {            int mid=(l+r)/2;            if(check(n,mid))            {                ans=mid;                l=mid+1;            }            else r=mid-1;        }        if(ans<4)        {            printf("0\n");        }        else        {            printf("%d\n",ans+1);        }    }    return 0;}