POJ 1743 Musical Theme 后缀数组

来源：互联网发布：冒泡排序优化编辑：程序博客网时间：2024/06/03 09:14

题目：http://poj.org/problem?id=1743

题意：给定一个数组，求不重叠的最长相同变化的长度

思路：后缀数组第一题。看后缀数组看了一天，整个人都懵逼了，找了一道基础的后缀数组题切一下。。。这道题是利用height数组的性质，height数组代表排名为i和i-1的后缀子串的最长公共前缀长度，公共前缀即是相同的变化。二分枚举答案，判断height数组大于等于枚举值的两起点间的距离是否大于等于k，若满足则两段不重复。

总结：这道题数据很弱，，，对于一组数据得出两种答案，交上去都能过，，，

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define debug() puts("here")using namespace std;const int N = 100010;int wa[N], wb[N], wv[N], wt[N], sa[N];int rnk[N], height[N];int n, arr[N];bool 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;    for(i = 0; i < m; i++) wt[i] = 0;    for(i = 0; i < n; i++) wt[x[i]=r[i]]++;    for(i = 1; i < m; i++) wt[i] += wt[i-1];    for(i = n - 1; i >= 0; i--) sa[--wt[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++) wt[i] = 0;        for(i = 0; i < n; i++) wt[wv[i]]++;        for(i = 1; i < m; i++) wt[i] += wt[i-1];        for(i = n - 1; i >= 0; i--) sa[--wt[wv[i]]] = y[i];        for(swap(x, y), 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 getheight(int *r, int *sa, int n){    int i, j, k = 0;    for(i = 0; i <= n; i++) rnk[sa[i]] = i;    for(i = 0; i < n; height[rnk[i++]] = k)        for(k ? k-- : 0, j = sa[rnk[i]-1]; r[i+k] == r[j+k]; k++);    printf("%d\n", height[n]);}bool judge(int k){    int i = 2;    while(true)    {        while(i <= n && height[i] < k) i++;        if(i > n) break;        int maxx = sa[i-1], minn = sa[i-1];        while(i <= n && height[i] >= k)            maxx = max(maxx, sa[i]), minn = min(minn, sa[i]), i++;        if(maxx - minn >= k) return true;    }    return false;}int main(){    while(scanf("%d", &n), n)    {        for(int i = 0; i < n; i++)            scanf("%d", &arr[i]);        if(n < 10)        {            puts("0");            continue;        }        for(int i = 0; i < n - 1; i++)            arr[i] = arr[i+1] - arr[i] + 100;        arr[--n] = 0;        da(arr, sa, n + 1, 200);        getheight(arr, sa, n);        int l = 4, r = (n - 1) / 2, res = 0;        while(l <= r)        {            int mid = (l + r) >> 1;            if(judge(mid)) l = mid + 1, res = mid;            else r = mid - 1;        }        printf("%d\n", res < 4 ? 0 : res + 1);    }    return 0;}

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