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Bzoj4199:[NOI2015]品酒大会

时间:2018-02-21 23:35:07      阅读:230      评论:0      收藏:0      [点我收藏+]

标签:getchar   efi   names   pac   online   ref   problem   大会   get   

题面

Bzoj4199

Sol

后缀数组
显然的暴力就是求\(LCP\)+差分
\(40\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(3e5 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int a[_], n, s[_], rk[_], sa[_], height[_], tmp[_], t[_];
ll ans1[_], ans2[_];
int st[20][_], lg[_];
char ss[_];

IL int Cmp(RG int i, RG int j, RG int k){
    return tmp[i] == tmp[j] && i + k <= n && j + k <= n && tmp[i + k] == tmp[j + k];
}

IL void Suffix_Sort(){
    RG int m = 26;
    for(RG int i = 1; i <= n; ++i) ++t[rk[i] = s[i]];
    for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
    for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
    for(RG int k = 1; k <= n; k <<= 1){
        RG int l = 0;
        for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
        for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
        for(RG int i = 0; i <= m; ++i) t[i] = 0;
        for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
        for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
        for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
        swap(tmp, rk), rk[sa[1]] = l = 1;
        for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
        if(l >= n) break;
        m = l;
    }
    for(RG int i = 1, h = 0; i <= n; ++i){
        if(h) --h;
        while(s[i + h] == s[sa[rk[i] - 1] + h]) ++h;
        height[rk[i]] = h;
    }
}

IL void ST_Prepare(){
    for(RG int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
    for(RG int i = 1; i <= n; ++i) st[0][i] = height[i];
    for(RG int i = 1; i <= lg[n]; ++i)
        for(RG int j = 1; j + (1 << i) - 1 <= n; ++j)
            st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
}

IL int LCP(RG int x, RG int y){
    x = rk[x], y = rk[y];
    if(x > y) swap(x, y); ++x;
    RG int len = y - x + 1;
    return min(st[lg[len]][x], st[lg[len]][y - (1 << lg[len]) + 1]);
}

int main(RG int argc, RG char* argv[]){
    Fill(ans2, -127), n = Input(), scanf(" %s", ss + 1);
    for(RG int i = 1; i <= n; ++i) s[i] = ss[i] - 'a' + 1, a[i] = Input();
    Suffix_Sort(), ST_Prepare();
    for(RG int i = 1; i < n; ++i)
        for(RG int j = i + 1; j <= n; ++j){
            RG int lcp = LCP(i, j);
            ++ans1[0], --ans1[lcp + 1];
            ans2[lcp] = max(ans2[lcp], 1LL * a[i] * a[j]);
        }
    for(RG int i = 0; i < n; ++i) ans1[i] += ans1[i - 1];
    for(RG int i = n - 1; ~i; --i) ans2[i] = max(ans2[i], ans2[i + 1]);
    for(RG int i = 0; i < n; ++i) printf("%lld %lld\n", ans1[i], ans2[i] == ans2[n] ? 0 : ans2[i]);
    return 0;
}

正解,把\(height\)从大到小排序
每次合并\(height\)两边的集合,因为从大到小,所以两边集合的\(LCP\)一定就是这个\(height\),因为有负数权值,维护集合的最大值,最小值组合

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(3e5 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int a[_], n, s[_], rk[_], sa[_], height[_], tmp[_], t[_];
ll ans1[_], ans2[_];
int fa[_], id[_], size[_], mx[_], mn[_];
char ss[_];

IL int Cmp(RG int i, RG int j, RG int k){
    return tmp[i] == tmp[j] && i + k <= n && j + k <= n && tmp[i + k] == tmp[j + k];
}

IL int _Cmp(RG int x, RG int y){
    return height[x] > height[y];
}

IL void Suffix_Sort(){
    RG int m = 26;
    for(RG int i = 1; i <= n; ++i) ++t[rk[i] = s[i]];
    for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
    for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
    for(RG int k = 1; k <= n; k <<= 1){
        RG int l = 0;
        for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
        for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
        for(RG int i = 0; i <= m; ++i) t[i] = 0;
        for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
        for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
        for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
        swap(tmp, rk), rk[sa[1]] = l = 1;
        for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
        if(l >= n) break;
        m = l;
    }
    for(RG int i = 1, h = 0; i <= n; ++i){
        if(h) --h;
        while(s[i + h] == s[sa[rk[i] - 1] + h]) ++h;
        height[rk[i]] = h;
    }
}

IL int Find(RG int x){
    return fa[x] == x ? x : fa[x] = Find(fa[x]);
}

int main(RG int argc, RG char* argv[]){
    Fill(ans2, -127), n = Input(), scanf(" %s", ss + 1);
    for(RG int i = 1; i <= n; ++i) s[i] = ss[i] - 'a' + 1, a[i] = Input();
    for(RG int i = 1; i <= n; ++i) fa[i] = id[i] = i, size[i] = 1, mx[i] = mn[i] = a[i];
    Suffix_Sort(), sort(id + 2, id + n + 1, _Cmp);
    for(RG int i = 2; i <= n; ++i){
        RG int fx = Find(sa[id[i] - 1]), fy = Find(sa[id[i]]);
        ans1[height[id[i]]] += 1LL * size[fx] * size[fy];
        fa[fy] = fx, size[fx] += size[fy];
        ans2[height[id[i]]] = max(ans2[height[id[i]]], max(1LL * mx[fx] * mx[fy], 1LL * mx[fx] * mn[fy]));
        ans2[height[id[i]]] = max(ans2[height[id[i]]], max(1LL * mn[fx] * mx[fy], 1LL * mn[fx] * mn[fy]));
        mx[fx] = max(mx[fx], mx[fy]);
        mn[fx] = min(mn[fx], mn[fy]);
    }
    for(RG int i = n - 1; ~i; --i) ans1[i] += ans1[i + 1], ans2[i] = max(ans2[i], ans2[i + 1]);
    for(RG int i = 0; i < n; ++i) printf("%lld %lld\n", ans1[i], ans2[i] == ans2[n] ? 0 : ans2[i]);
    return 0;
}

Bzoj4199:[NOI2015]品酒大会

标签:getchar   efi   names   pac   online   ref   problem   大会   get   

原文地址:https://www.cnblogs.com/cjoieryl/p/8457716.html

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