POJ 1743 不可重叠最长重复子串

#include <cstdio>using namespace std;//---------------------------------------------//功能:2倍增算法生成后缀数组,参考罗穗骞论文//sa:返回的后缀数组//str:输入的串(这里为int数组)//n:串长度//M:串中元素的最大值,最小值为0//---------------------------------------------#define MAX_N 19999void suffix_array(int* sa, int* str, int n, int M){static int pits[MAX_N];//挖坑计数数组static int aux[MAX_N];//用于计算sa时的辅助数组static int xx[MAX_N], yy[MAX_N];int* x = xx, *y = yy, *t;//x[]总保存当前的rankint i, j, r, p, higher1, higher2;//先对所有后缀的第一个字符进行排序(采用挖坑式的基数排序,即统计每个字符的个数,以便在扫描时总能将字符放入合适的位置),放入sa中for (i = 0; i <= M; ++i){pits[i] = 0;}for (i = 0; i < n; ++i){++pits[x[i] = str[i]];}for (i = 1; i <= M; ++i) //挖坑{pits[i] += pits[i - 1];}for (i = n - 1; i >= 0; --i){sa[--pits[str[i]]] = i;}//进行若干趟基数排序for (i = 1, r = M; i < n; i *= 2){//先计算出第一关键字的各个坑,以便进行基数排序for (j = 0; j <= r; ++j){pits[j] = 0;}for (j = 0; j < n; ++j){++pits[x[j]];}for (j = 1; j <= r; ++j){pits[j] += pits[j - 1];}//按第二关键字排序,排序之后放入aux数组中for (j = n - i, p = 0; j < n; ++j, ++p){aux[p] = j;}for (j = 0; j < n; ++j){if (sa[j] >= i){aux[p] = sa[j] - i;++p;}}//从后到前扫描aux数组,根据挖好的第一关键字的坑收集到sa数组中,完成一次基数排序for (j = n - 1; j >= 0; --j){sa[--pits[x[aux[j]]]] = aux[j];}//算出新的rank数组(即x[])for (t = x, x = y, y = t, x[sa[0]] = 0, r = 0, j = 1; j < n; ++j){higher1 = sa[j] + i;higher2 = sa[j - 1] + i;if (y[sa[j]] != y[sa[j - 1]]|| (!(higher1 >= n && higher2 >= n || higher1 < n && higher2 < n && y[higher1] == y[higher2]))){++r;}x[sa[j]] = r;}}}//功能:根据后缀数组生成rank数组void gen_rank(int* rank, int* sa, int n){int i;for (i = 0; i < n; ++i){rank[sa[i]] = i;}}//功能:根据串、后缀数组和排名数组计算height数组void calc_height(int* height, int* str, int* sa, int* rank, int n){//根据h[i]>=h[i-1]简化计算int i, j, k;height[0] = 0;for (k = 0, i = 0; i < n; height[rank[i]] = k, ++i){if (k){--k;}if (rank[i] >= 1){for (j = sa[rank[i] - 1]; j + k < n && i + k < n && str[i + k] == str[j + k]; ++k);}}}//功能:查找是否存在长度为k的不重叠重复子串bool judge(int* sa, int* rank, int* height, int n, int k){int mini, maxi, i;for (mini = maxi = sa[0], i = 1; i < n; ++i){if (height[i] < k){mini = maxi = sa[i];}else if (sa[i] < mini){mini = sa[i];if (maxi - mini >= k){return true;}}else if (sa[i] > maxi){maxi = sa[i];if (maxi - mini >= k){return true;}}}return false;}int notes[20000], sa[19999], rank[19999], height[19999];int main(){int N, i, low, high, mid;for (; ;){//输入scanf("%d", &N);if (!N){break;}for (i = 0; i < N; ++i){scanf("%d", &notes[i]);}for (i = N - 1; i >= 0; --i){notes[i] -= notes[i - 1];notes[i] += 87;}//生成后缀数组suffix_array(sa, notes + 1, N - 1, 174);//生成rank数组gen_rank(rank, sa, N - 1);//算出height数组calc_height(height, notes + 1, sa, rank, N - 1);//二分寻找klow = 0;high = (N - 1) / 2;while (low < high){mid = (low + high + 1) / 2;if (judge(sa, rank, height, N - 1, mid)){low = mid;}else{high = mid - 1;}}if (low < 4){puts("0");}else{printf("%d/n", low + 1);}}return 0;}

 

几个输入数据:

10
1 2 3 4 5 6 7 8 9 10
8
1 1 2 1 1 1 1 2
30
25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18 82 78 74 70 66 67
64 60 65 80
0

 

我用这个模板提交了2774,不过速度很慢,G++下735ms:

#include <cstring>#include <cstdio>using namespace std;#define MAX_N 200002void suffix_array(int* sa, char* str, int n, int M){static int pits[MAX_N];//挖坑计数数组static int aux[MAX_N];//用于计算sa时的辅助数组static int xx[MAX_N], yy[MAX_N];int* x = xx, *y = yy, *t;//x[]总保存当前的rankint i, j, r, p, higher1, higher2;//先对所有后缀的第一个字符进行排序(采用挖坑式的基数排序,即统计每个字符的个数,以便在扫描时总能将字符放入合适的位置),放入sa中for (i = 0; i <= M; ++i){pits[i] = 0;}for (i = 0; i < n; ++i){++pits[x[i] = str[i]];}for (i = 1; i <= M; ++i) //挖坑{pits[i] += pits[i - 1];}for (i = n - 1; i >= 0; --i){sa[--pits[str[i]]] = i;}//进行若干趟基数排序for (i = 1, r = M; i < n; i *= 2){//先计算出第一关键字的各个坑,以便进行基数排序for (j = 0; j <= r; ++j){pits[j] = 0;}for (j = 0; j < n; ++j){++pits[x[j]];}for (j = 1; j <= r; ++j){pits[j] += pits[j - 1];}//按第二关键字排序,排序之后放入aux数组中for (j = n - i, p = 0; j < n; ++j, ++p){aux[p] = j;}for (j = 0; j < n; ++j){if (sa[j] >= i){aux[p] = sa[j] - i;++p;}}//从后到前扫描aux数组,根据挖好的第一关键字的坑收集到sa数组中,完成一次基数排序for (j = n - 1; j >= 0; --j){sa[--pits[x[aux[j]]]] = aux[j];}//算出新的rank数组(即x[])for (t = x, x = y, y = t, x[sa[0]] = 0, r = 0, j = 1; j < n; ++j){higher1 = sa[j] + i;higher2 = sa[j - 1] + i;if (y[sa[j]] != y[sa[j - 1]]|| (!(higher1 >= n && higher2 >= n || higher1 < n && higher2 < n && y[higher1] == y[higher2]))){++r;}x[sa[j]] = r;}}}//功能:根据后缀数组生成rank数组void gen_rank(int* rank, int* sa, int n){int i;for (i = 0; i < n; ++i){rank[sa[i]] = i;}}//功能:根据串、后缀数组和排名数组计算height数组void calc_height(int* height, char* str, int* sa, int* rank, int n){//根据h[i]>=h[i-1]简化计算int i, j, k;height[0] = 0;for (k = 0, i = 0; i < n; height[rank[i]] = k, ++i){if (k){--k;}if (rank[i] >= 1){for (j = sa[rank[i] - 1]; j + k < n && i + k < n && str[i + k] == str[j + k]; ++k);}}}char str[200001];int sa[200001], rank[200001], height[200001];int main(){int l1, l2, l, i, maxi;scanf("%s", str);l1 = strlen(str);str[l1] = 'z' + 1;scanf("%s", str + l1 + 1);l2 = strlen(str + l1 + 1);l = l1 + l2 + 1;for (i = 0; i < l; ++i){str[i] -= 'a';}suffix_array(sa, str, l, 26);gen_rank(rank, sa, l);calc_height(height, str, sa, rank, l);for (maxi = 0, i = 1; i < l; ++i){if (height[i] > maxi && (sa[i - 1] < l1 && sa[i] > l1 || sa[i - 1] > l1 && sa[i] < l1)){maxi = height[i];}}printf("%d/n", maxi);return 0;}