[POJ1743]Musical Theme[SA+二分]
2009那篇论文里介绍的做法
这里面有下载链接
这个题是要转成差分序列再做的(同时要对答案处理一下)
导致我WA了一发的地方 if (Max - Min > x) return 1;
写成>=了(求最长不重叠重复子串确实是>=)
对于下面这两组数据
5
1 1 1 1 1
6
1 1 1 1 1 1
这两组数据在差分序列上的最长不重叠重复子串长度都是 2 ,但第一组的实际答案是 2 ,第二组是 3 ,所以要使得第一组二分出来的值是 1 (我输出的是二分的答案加 1 )
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int k, n, rnk[20005], sa[20005], s[20005], H[20005], c[20005];
void Sort(int *x, int *y, int *rk) {
static int C[20005];
for (int i = 0; i <= k; ++i) C[i] = 0;
for (int i = 1; i <= n; ++i) ++C[rk[i]];
for (int i = 1; i <= k; ++i) C[i] += C[i - 1];
for (int i = n; i; --i) y[C[rk[x[i]]]--] = x[i];
}
inline bool cmp(int *y, int a, int b, int m) {return y[a] == y[b] && y[a + m] == y[b + m];}
void get_SA() {
static int Y[20005];
int *y = Y, *rk = rnk;
k = 200;
for (int i = 1; i <= n; ++i) rk[i] = s[y[i] = i];
Sort(y, sa, rk);
for (int m = 1, p = 0; p < n; k = p, m <<= 1) {
for (p = 0; p < m; ++p) y[p + 1] = n - m + p + 1;
for (int i = 1; i <= n; ++i)
if (sa[i] > m) y[++p] = sa[i] - m;
Sort(y, sa, rk), swap(y, rk);
rk[sa[p = 1]] = 1;
for (int i = 2; i <= n; ++i) rk[sa[i]] = cmp(y, sa[i - 1], sa[i], m) ? p : ++p;
}
for (int i = 1; i <= n; ++i) rnk[sa[i]] = i;
}
void get_H() {
for (int i = 1, k = 0; i <= n; H[rnk[i++]] = k)
for (k ? --k : 0; s[i + k] == s[sa[rnk[i] - 1] + k]; ++k);
}
bool check(int x) {
for (int i = 2, Min = sa[1], Max = sa[1]; i <= n; ++i) {
if (H[i] >= x) Min = min(Min, sa[i]), Max = max(Max, sa[i]);
else Max = Min = sa[i];
if (Max - Min > x) return 1;
}
return 0;
}
int main() {
while (~scanf("%d", &n) && n) {
for (int i = 1; i <= n; ++i) scanf("%d", c + i), s[i] = c[i] - c[i - 1] + 90;
get_SA();
get_H();
int l = 1, r = n, mid;
while (l <= r) {
if (check(mid = l + r >> 1)) l = mid + 1;
else r = mid - 1;
}
cout << (l < 5 ? 0 : l) << endl;//l-1<4?0:l-1+1
}
return 0;
}