BZOJ 4278: [ONTAK2015]Tasowanie (后缀数组 / 二分+hash)

直接归并,然后如果哪边的后缀字典序比较小就去哪边,然后就可以后缀数组 博客传送门…

但是本蒟蒻不会后缀数组
$Upd:$现在会了233.一道差不多的题:BZOJ 1692: [Usaco2007 Dec]队列变换

于是就hash后二分找相同的最长区间,然后比较后一个谁更小…

时间复杂度$O(nlogn)$.

注意先在最后加一个极大值.因为如果一个序列A到末尾刚好与序列B的一段相等,那么一定是取B更优,因为B后面可能有更小的.

hack数据

4 2 2 2 1
3 2 2 2

如果没有加极大值,OJ上能A,但是这组数据会WA掉.

CODE

#include<bits/stdc++.h>
using namespace std;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
	char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg=-flg;
	for(res=ch-'0';isdigit(ch=getchar());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 200005 ;
const int p = 137;
int n, m, a[MAXN], b[MAXN], A[MAXN], B[MAXN], mul[MAXN];
inline int hsh(int *H, int l, int r) {
    return l <= r ? H[r] - H[l-1] * mul[r-l+1] : 0;
}
inline int lcp(int i, int j) {
    int l = 0, r = min(n-i+1, m-j+1), mid;
    while(l < r) {
        mid = (l + r + 1) >> 1;
        if(hsh(A, i, i+mid-1) == hsh(B, j, j+mid-1)) l = mid;
        else r = mid-1;
    }
    return l;
}
int main() {
	int i, j, k;
	read(n); for(i = 1; i <= n; ++i) read(a[i]), A[i] = A[i-1] * p + a[i]; a[n+1] = 1005;
    read(m); for(i = 1; i <= m; ++i) read(b[i]), B[i] = B[i-1] * p + b[i]; b[m+1] = 1005;
    for(mul[0] = 1, i = 1; i <= n || i <= m; ++i) mul[i] = mul[i-1] * p;
    for(i = j = 1; i <= n && j <= m; printf("%d ", a[i+k] < b[j+k] ? a[i++] : b[j++])) k = lcp(i, j);
    while(i <= n) printf("%d ", a[i++]);
    while(j <= m) printf("%d ", b[j++]);
}