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【POJ】1743 Musical Theme

http://poj.org/problem?id=1743

题意:不可重叠最长重复子串,n<=20000,具体看《后缀数组》-- 罗穗骞

#include <cstdio>#include <algorithm>using namespace std;const int N=20015;void sort(int *x, int *y, int *sa, int n, int m) {	static int c[N], i;	for(i=0; i<m; ++i) c[i]=0;	for(i=0; i<n; ++i) c[x[y[i]]]++;	for(i=1; i<m; ++i) c[i]+=c[i-1];	for(i=n-1; i>=0; --i) sa[--c[x[y[i]]]]=y[i];}void hz(int *r, int *sa, int n, int m) {	static int t1[N], t2[N];	static int *x, *y, *t, j, i, p=0;	x=t1; y=t2;	for(i=0; i<n; ++i) x[i]=r[i], y[i]=i;	sort(x, y, sa, n, m);	for(j=1, p=1; p<n; j<<=1, m=p) {		p=0;		for(i=n-j; i<n; ++i) y[p++]=i;		for(i=0; i<n; ++i) if(sa[i]-j>=0) y[p++]=sa[i]-j;		sort(x, y, sa, n, m);		for(t=x, x=y, y=t, x[sa[0]]=0, p=1, 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++;	}}void geth(int *a, int *sa, int *rank, int *h, int n) {	static int k, i, j; k=0;	for(i=1; i<=n; ++i) rank[sa[i]]=i;	for(i=1; i<=n; h[rank[i++]]=k)		for(k?--k:0, j=sa[rank[i]-1]; a[i+k]==a[j+k]; ++k);}const int oo=~0u>>2;int sa[N], rank[N], h[N], n, a[N];bool check(int k) {	int mx=sa[1], mn=sa[1];	for(int i=2; i<=n; ++i) {		if(h[i]>=k) {			mx=max(mx, sa[i]);			mn=min(mn, sa[i]);			if(mx-mn>=k) return 1;		}		else mx=sa[i], mn=sa[i];	}	return 0;}int main() {	while(scanf("%d", &n), n) {		scanf("%d", &a[1]); --n;		for(int i=1; i<=n; ++i) scanf("%d", &a[i+1]), a[i]=a[i+1]-a[i]+100;		hz(a, sa, n+1, 200);		geth(a, sa, rank, h, n);		int mid, l=0, r=n/2;		while(l<=r) {			mid=(l+r)>>1;			if(check(mid)) l=mid+1;			else r=mid-1;		}		if(l>=5) printf("%d\n", l);		else puts("0");	}	return 0;}

  


 

经典题....我们求出height数组后,按二分的大小k分组。即每个组里的height值都要>=k,然后看每个块是否有sa值之差>=k的即可

(吐槽:poj不能用bits/stdc++.h啊啊啊啊啊啊ce了两发啊...

【POJ】1743 Musical Theme