BZOJ3238 [Ahoi2013]差异 【后缀数组 + 单调栈】

题目链接

BZOJ3238

题解

简单题
经典后缀数组 + 单调栈套路,求所有后缀\(lcp\)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define cls(s) memset(s,0,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
char s[maxn];
int sa[maxn],rank[maxn],height[maxn],bac[maxn],t1[maxn],t2[maxn],n,m;
void getsa(){
    int *x = t1,*y = t2; m = 255;
    for (int i = 0; i <= m; i++) bac[i] = 0;
    for (int i = 1; i <= n; i++) bac[x[i] = s[i]]++;
    for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
    for (int i = n; i; i--) sa[bac[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 > 0) y[++p] = sa[i] - k;
        for (int i = 0; i <= m; i++) bac[i] = 0;
        for (int i = 1; i <= n; i++) bac[x[y[i]]]++;
        for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
        for (int i = n; i; i--) sa[bac[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;
    for (int i = 1,k = 0; i <= n; i++){
        if (k) k--;
        int j = sa[rank[i] - 1];
        while (s[i + k] == s[j + k]) k++;
        height[rank[i]] = k;
    }
}
cp st[maxn],t;
int top;
void solve(){
    LL sum = 0,ans = 0;
    for (int i = 2; i <= n; i++){
        t = mp(height[i],1);
        while (top && st[top].first >= t.first){
            sum -= 1ll * st[top].first * st[top].second;
            t.second += st[top].second;
            top--;
        }
        st[++top] = t;
        sum += 1ll * t.first * t.second;
        ans += sum;
    }
    printf("%lld\n",1ll * n * (n + 1) * (n - 1) / 2 - 2 * ans);
}
int main(){
    scanf("%s",s + 1); n = strlen(s + 1);
    getsa();
    solve();
    return 0;
}

猜你喜欢

转载自www.cnblogs.com/Mychael/p/9022809.html