[NOI2015]品酒大会

建出SAM和后缀树

两个子串的最长lcp就是他们的lca

因此我们只需求出最长的个数然后用前缀和就能算出总个数，最大值也是同样的方法， 当然要逆序建SAM才能保证他们的第一位相同。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define LL long long

using namespace std;

inline LL read()
{
    LL x = 0, w = 1; char ch = 0;
    while(ch < '0' || ch > '9') {
        if(ch == '-') {
            w = -1;
        }
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * w;
}

const int MAXN = 1e6 + 10;

int head[MAXN];
char s[MAXN];
LL d[MAXN];
LL ans[MAXN], num[MAXN];
LL fi, se;
int flag[MAXN];
LL size[MAXN];
LL maxn[MAXN], minn[MAXN];
int len;
int lst = 1, tot = 1;

struct SAM {
    int c[27];
    int fa, len;
} t[MAXN];

void extend(int x, int d)
{
    int last = lst, np = ++tot;
    t[np].len = t[last].len + 1;
    size[np] = 1;
    lst = np;
    maxn[np] = minn[np] = d;
    while(last && !t[last].c[x]) {
        t[last].c[x] = np;
        last = t[last].fa; 
    }
    if(!last) {
        t[np].fa = 1;
    } else {
        int q = t[last].c[x];
        if(t[q].len == t[last].len + 1) {
            t[np].fa = q;
        } else {
            int nq = ++tot;
            flag[nq] = 1;
            maxn[nq] = -2e18, minn[nq] = 2e18;
            t[nq] = t[q];
            t[nq].len = t[last].len + 1;
            t[np].fa = t[q].fa = nq;
            while(t[last].c[x] == q) {
                t[last].c[x] = nq;
                last = t[last].fa;
            }
        }
    }
}

int cnt = 0;

struct edge {
    int v, next;
} g[MAXN];

void addedge(int u, int v)
{
    g[++cnt].v = v;
    g[cnt].next = head[u];
    head[u] = cnt;    
}

bool ok(int x)
{
    return maxn[x] != -2e18 && minn[x] != 2e18;
}

void DFS(int x)
{    
    for(int j = head[x]; j; j = g[j].next) {
        int to = g[j].v;
        DFS(to);
        num[t[x].len] += size[x] * size[to];
        size[x] += size[to];
        if(ok(x) && ok(to)) {
            ans[t[x].len] = max(ans[t[x].len], max(maxn[x] * maxn[to], minn[x] * minn[to]));
        }
        maxn[x] = max(maxn[x], maxn[to]);
        minn[x] = min(minn[x], minn[to]);
    }
}

int main()
{
    //freopen("drink.out", "w", stdout);
    len = read();
    scanf("%s", s + 1);
    fi = -2e18, se = 2e18;
    maxn[1] = -2e18, minn[1] = 2e18;
    for(int i = 1; i <= len; i++) {
        d[i] = read();
        if(fi != -2e18)
        ans[0] = max(fi * d[i], ans[0]);
        if(se != 2e18)
        ans[0] = max(se * d[i], ans[0]);
        fi = max(fi, d[i]), se = min(se, d[i]);
    }
    for(int i = 0; i <= len; i++) {
        ans[i] = -2e18;
    }
    for(int i = len; i >= 1; i--) {
        extend(s[i] - 'a', d[i]);
    }
    for(int i = 2; i <= tot; i++) {
        addedge(t[i].fa, i);
    }
    DFS(1);
    for(int i = len - 1; i >= 0; i--) {
        num[i] += num[i + 1];
        ans[i] = max(ans[i], ans[i + 1]);
    }
    for(int i = 0; i < len; i++) {
        if(num[i] == 0) {
            printf("0 0
");
        } else {
            printf("%lld %lld
", num[i], ans[i]);
        }
    }
    return 0;
}