POJ 3261 Milk Patterns[后缀数组]
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题目链接:http://poj.org/problem?id=3261
题意:N头牛,每头牛有一个编号,求可重叠的“至少”出现K次的“最大子串”。
思路:后缀数组+二分。
代码:
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int nMax =1000012;int num[nMax];int sa[nMax], rank[nMax], height[nMax];int wa[nMax], wb[nMax], wv[nMax], wd[nMax];int 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 n, int m){ // 倍增算法 r为待匹配数组 n为总长度 m为字符范围 int i, j, p, *x = wa, *y = wb, *t; for(i = 0; i < m; i ++) wd[i] = 0; for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++; for(i = 1; i < m; i ++) wd[i] += wd[i-1]; for(i = n-1; i >= 0; i --) sa[-- wd[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 ++) wd[i] = 0; for(i = 0; i < n; i ++) wd[wv[i]] ++; for(i = 1; i < m; i ++) wd[i] += wd[i-1]; for(i = n-1; i >= 0; i --) sa[-- wd[wv[i]]] = y[i]; for(t = x, x = y, y = t, 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 calHeight(int *r, int n){ // 求height数组。 int i, j, k = 0; for(i = 1; i <= n; i ++) rank[sa[i]] = i; for(i = 0; i < n; height[rank[i ++]] = k){ for(k ? k -- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k ++); }}bool check(int mid,int len,int k){ //长度为mid的子串是否出现了k次 int i,j,tot=0; for(i=2;i<=len;i++){ if(height[i]<mid) tot=0; else{ tot++; if(tot==k-1) return 1; } } return 0;}int main(){ //freopen("C:\\Users\\acm\\Desktop\\001.in","r",stdin); int i,j,n,ans,k; while(scanf("%d%d",&n,&k)!=EOF){ ans=0; for(i=0;i<n;i++){ scanf("%d",&num[i]); num[i]++; } num[n]=0; da(num,n+1,1000002); calHeight(num,n); int left=0,right=n,mid; while(right>=left){ mid=(right+left)/2; if(check(mid,n,k)){ left=mid+1; ans=mid; } else{ right=mid-1; } } printf("%d\n",ans); } return 0;}
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