POJ1743 Musical Theme 题解&代码

来源：互联网发布：开淘宝网店怎么推广编辑：程序博客网时间：2024/05/11 00:15

后缀数组…
对最长不重叠子串长度进行二分判定，判定方式是暴力分组
利用height[]的性质：如果height[i]>x,height[i-1]>x，那么存在从sa[i-2]到sa[i]的部分最长公共子串大于x
可以推论：对于height[i]~height[j]均大于x，那么存在sa[i-1]到sa[j]的部分最长公共子串大于x
这样可以得出对于一段连续的满足条件的height[]，有连续的多个sa[]公共子串满足条件，用l和r维护最长起始位置差
That’s all.

#include <iostream>#include <cstdio>using namespace std;const int maxn = 20005;int n, l, r, mid, str[maxn], s[maxn];int wa[maxn], wb[maxn], wv[maxn], cnt[maxn];int sa[maxn], height[maxn], Rank[maxn];void DA(int *r,int n,int m){        int *x = wa, *y = wb, p;        for(int i = 0; i < m; i++) cnt[i] = 0;        for(int i = 0; i < n; i++) cnt[ x[i] = r[i] ]++;        for(int i = 1; i < m; i++) cnt[i] += cnt[i-1];        for(int i = n-1; i >= 0; i--) sa[--cnt[x[i]]] = i;        for(int j = 1; j < n ; j=j<<1)        {            p = 0;            for(int i = n-j; i < n; i++) y[p++] = i;            for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;            for(int i = 0; i < n; i++) wv[i] = x[y[i]];            for(int i = 0; i < m; i++) cnt[i] = 0;            for(int i = 0; i < n; i++) cnt[wv[i]]++;            for(int i = 1; i < m; i++) cnt[i] += cnt[i-1];            for(int i = n-1; i >= 0; i--) sa[--cnt[wv[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]] == y[sa[i-1]]) && (y[sa[i]+j] == y[sa[i-1]+j]) ? p-1 : p++;        if(p >= n) break;        m = p;    }}void calheight(int *r,int n){    for(int i = 1; i <= n; i++)Rank[sa[i]] = i;    int j = 0, k = 0;    for(int i = 0; i < n; height[Rank[i]] = k, i++)        for(k?k--:k=0, j = sa[Rank[i]-1]; r[i+k] == r[j+k]; k++);}bool check(int x){    int l = sa[0], r = sa[0];    for(int i = 0; i < n; i++)    {        if(height[i] < x)        {            l = sa[i];            r = sa[i];            continue;        }        l = min(l, sa[i]);        r = max(r, sa[i]);        if(r-l >= x) return true;    }    return false;}int main(void){    while(scanf("%d", &n) && n)    {        for(int i = 0; i < n; i++)        {            scanf("%d", &str[i]);            if(i) s[i-1] = str[i]-str[i-1], s[i-1] += 100;        }        s[n-1] = 0;        DA(s, n, 200);        calheight(s, n-1);        l = 0, r = n-1;        while(l < r)        {            mid = (l+r)/2;            if(check(mid)) l = mid+1;            else r = mid;        }        printf("%d\n",l >= 5? l : 0 );    }    return 0;}

0 0