[POJ1743]Musical Theme
试题描述
A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings.
Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it:
- is at least five notes long
- appears (potentially transposed -- see below) again somewhere else in the piece of music
- is disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)
Transposed means that a constant positive or negative value is added to every note value in the theme subsequence.
Given a melody, compute the length (number of notes) of the longest theme.
One second time limit for this problem's solutions!
输入
The last test case is followed by one zero.
输出
For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.
输入示例
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
输出示例
5
数据规模及约定
见“试题描述”
题解
差分后二分 + 后缀数组。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 20010
int n, S[maxn], rank[maxn], height[maxn], sa[maxn], Ws[maxn];
bool cmp(int* a, int p1, int p2, int l) {
if(p1 + l > n && p2 + l > n) return a[p1] == a[p2];
if(p1 + l > n || p2 + l > n) return 0;
return a[p1] == a[p2] && a[p1+l] == a[p2+l];
}
void ssort() {
int *x = rank, *y = height;
int m = 0;
memset(Ws, 0, sizeof(Ws));
for(int i = 1; i <= n; i++) Ws[x[i] = S[i]]++, m = max(m, x[i]);
for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
for(int i = n; i; i--) sa[Ws[x[i]]--] = i;
for(int j = 1, pos = 0; pos < n; j <<= 1, m = pos) {
pos = 0;
for(int i = n - j + 1; i <= n; i++) y[++pos] = i;
for(int i = 1; i <= n; i++) if(sa[i] > j) y[++pos] = sa[i] - j;
for(int i = 1; i <= m; i++) Ws[i] = 0;
for(int i = 1; i <= n; i++) Ws[x[i]]++;
for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
for(int i = n; i; i--) sa[Ws[x[y[i]]]--] = y[i];
swap(x, y); pos = 1; x[sa[1]] = 1;
for(int i = 2; i <= n; i++) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? pos : ++pos;
}
return ;
}
void calch() {
for(int i = 1; i <= n; i++) rank[sa[i]] = i;
for(int i = 1, j, k = 0; i <= n; height[rank[i++]] = k)
for(k ? k-- : 0, j = sa[rank[i]-1]; S[j+k] == S[i+k]; k++);
return ;
}
bool check(int x) {
int mn = sa[1], mx = sa[1];
for(int i = 2; i <= n; i++)
if(height[i] < x) mn = mx = sa[i];
else {
mn = min(mn, sa[i]); mx = max(mx, sa[i]);
if(mx - mn >= x + 1) return 1;
}
return 0;
}
int main() {
while(1) {
n = read();
if(!n) break;
for(int i = 1; i <= n; i++) S[i] = read();
for(int i = 1; i <= n; i++) S[i] = S[i+1] - S[i] + 88;
n--;
ssort();
calch();
int l = 0, r = n + 1;
while(r - l > 1) {
int mid = l + r >> 1;
if(check(mid)) l = mid; else r = mid;
}
l++;
if(l >= 5) printf("%d\n", l);
else puts("0");
}
return 0;
}
后缀自动机建出来后在 parent 树上 dp 一下，每个节点 i 记录 mxr[i] 和 mnr[i] 分别表示该节点最大和最小的 right 值，那么最后这个节点的答案就是 min(mxr[i] - mnr[i], Max[i])。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 40010
#define maxa 175
#define oo 2147483647
int n, S[maxn];
int rt, last, ToT, to[maxn][maxa], par[maxn], Max[maxn], rgt[maxn];
void extend(int x, int pos) {
int p = last, np = ++ToT; Max[np] = Max[p] + 1; last = np; rgt[np] = pos;
memset(to[np], 0, sizeof(to[np]));
while(p && !to[p][x]) to[p][x] = np, p = par[p];
if(!p){ par[np] = rt; return ; }
int q = to[p][x];
if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
int nq = ++ToT; Max[nq] = Max[p] + 1; rgt[nq] = 0;
memcpy(to[nq], to[q], sizeof(to[q]));
par[nq] = par[q];
par[q] = par[np] = nq;
while(p && to[p][x] == q) to[p][x] = nq, p = par[p];
return ;
}
int sa[maxn], Ws[maxn], mxr[maxn], mnr[maxn];
void build() {
memset(Ws, 0, sizeof(Ws));
for(int i = 1; i <= ToT; i++) Ws[n-Max[i]]++;
for(int i = 1; i <= n; i++) Ws[i] += Ws[i-1];
for(int i = ToT; i; i--) sa[Ws[n-Max[i]]--] = i;
for(int i = 1; i <= ToT; i++)
if(!rgt[i]) mxr[i] = -1, mnr[i] = oo;
else mxr[i] = mnr[i] = rgt[i];
return ;
}
int main() {
while(1) {
n = read(); if(!n) break;
for(int i = 1; i <= n; i++) S[i] = read(); n--;
for(int i = 1; i <= n; i++) S[i] = S[i+1] - S[i] + 87;
// for(int i = 1; i <= n; i++) printf("%d%c", S[i], i < n ? ' ' : '\n');
rt = last = ToT = 1; memset(to[1], 0, sizeof(to[1]));
for(int i = 1; i <= n; i++) extend(S[i], i);
// for(int i = 1; i <= ToT; i++) printf("%d%c", par[i], i < ToT ? ' ' : '\n');
build();
int ans = 0;
for(int i = 1; i <= ToT; i++) {
int u = sa[i];
mxr[par[u]] = max(mxr[par[u]], mxr[u]);
mnr[par[u]] = min(mnr[par[u]], mnr[u]);
ans = max(ans, min(mxr[u] - mnr[u] - 1, Max[u]));
}
ans++;
printf("%d\n", ans >= 5 ? ans : 0);
}
return 0;
}