【bzoj3238】[Ahoi2013]差异

3238: [Ahoi2013]差异

Time Limit: 20 Sec  Memory Limit: 512 MB
Submit: 3425  Solved: 1559
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Description

Input

一行，一个字符串S

Output

 

一行，一个整数，表示所求值

Sample Input

cacao

Sample Output

54

HINT

2<=N<=500000,S由小写英文字母组成

Source

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这题很尬。。。。。。

我能说我是看错题目才想到这种做法的吗。。？

首先对于那个和式，很容易发现和按sa数组下标的顺序算的答案是一样的

因为把两个式子乘二都等于算所有( i , j )有序対的差异

于是。。我们可以用后缀数组瞎搞。。。

建一下后缀数组，然后从后往前做，维护height的最小值的和即可

单调栈就可以了。。。然而蒟蒻很蠢居然还上了一棵线段树。。。。。。

PS.听说出题的目的是考后缀自动机。。？

代码：

#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int INF = 2147483647;const int maxn = 500100;const int maxseg = 4 * maxn;int m;char s[maxn];int rank[maxn],t1[maxn],t2[maxn],c[maxn];LL siz[maxseg],sum[maxseg],ans,n,sa[maxn],sumtj,height[maxn];bool flag[maxseg];inline void del(int o){flag[o] = 1; siz[o] = sum[o] = 0;}inline void pushdown(int o){if (flag[o]){int lc = o * 2,rc = o * 2 + 1;del(lc); del(rc);flag[o] = 0;}}inline void maintain(int o){int lc = o * 2,rc = o * 2 + 1;siz[o] = siz[lc] + siz[rc];sum[o] = sum[lc] + sum[rc];}inline void reset(int o,int l,int r,int al,int ar){if (al <= l && r <= ar){del(o); return;}pushdown(o);int lc = o * 2,rc = o * 2 + 1,mid = l + r >> 1;if (ar <= mid) reset(lc,l,mid,al,ar);if (mid < al) reset(rc,mid + 1,r,al,ar);if (al <= mid && mid < ar){reset(lc,l,mid,al,mid);reset(rc,mid + 1,r,mid + 1,ar);}maintain(o);}inline LL query(int o,int l,int r,int al,int ar){if (al <= l && r <= ar) return siz[o];pushdown(o);int lc = o * 2,rc = o * 2 + 1,mid = l + r >> 1;if (ar <= mid) return query(lc,l,mid,al,ar);if (mid < al) return query(rc,mid + 1,r,al,ar);if (al <= mid && mid < ar) return query(lc,l,mid,al,mid) + query(rc,mid + 1,r,mid + 1,ar);}inline void inc(int o,int l,int r,int pos,LL cnt){if (l == r){sum[o] += cnt * l; siz[o] += cnt; return;}pushdown(o);int mid = l + r >> 1,lc = o * 2,rc = o * 2 + 1;if (pos <= mid) inc(lc,l,mid,pos,cnt);else inc(rc,mid + 1,r,pos,cnt);maintain(o);}inline void build(){int *x = t1,*y = t2;for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[i] = s[i]]++;for (int i = 1; i <= m; i++) c[i] += c[i - 1];for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;for (int k = 1; k <= n; k <<= 1){int p = 0;for (int i = n - k + 1; i <= n; i++) y[++p] = i;for (int i = 1; i <= n; i++) if (sa[i] > k) y[++p] = sa[i] - k;for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[y[i]]]++;for (int i = 1; i <= m; i++) c[i] += c[i - 1];for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];swap(x,y);x[sa[1]] = p = 1;for (int i = 2; i <= n; i++)x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p : ++p;if (p >= n) break;m = p;}for (int i = 1; i <= n; i++) rank[sa[i]] = i;int k = 0;for (int i = 1; i <= n; i++){if (k) k--;int j = sa[rank[i] - 1];while (s[i + k] == s[j + k]) k++;height[rank[i]] = k;}}int main(){#ifdef dyffreopen("hyj1.txt","r",stdin);freopen("hyj2.txt","w",stdout);#endifscanf("%s",s + 1); n = strlen(s + 1);for (int i = 1; i <= n; i++) m = max(m,(int)s[i]);build();ans = 0;for (LL i = n - 1; i >= 1; i--){sumtj += n - sa[i + 1] + 1;ans += (n - sa[i] + 1) * (n - i) + sumtj;LL k = query(1,0,n,height[i + 1] + 1,n);reset(1,0,n,height[i + 1] + 1,n);inc(1,0,n,height[i + 1],k + 1);ans -= 2 * sum[1];}printf("%lld",ans);return 0;}