Codeforces Round #271 (Div. 2) E. Pillars

题目链接: http://codeforces.com/problemset/problem/474/E

题解: 设f[i]为以位置i为结束的字串，那么方程为f[i] = max{ f[j] } + 1 (abs(a[j] - a[i]) >= k)  复杂度O(n * n)，超时。

所有出现过的数字排序去重离散化，建线段树维护区间最大值，对于每个a[i] 查询[0, a[i] - k]  ∪ [a[i] + k, INF]的最大值，然后向线段树插入f[i]，总复杂度O(n * log(n))

代码:

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define LL long long#define N 400000using namespace std;struct node{    LL a, b;    node *l, *r;    LL s;    LL pos;}f[N], *top;LL cnt;LL num[N], a[N];LL dp[N];LL pre[N];LL n, d;LL q[N];node* build(LL a, LL b){    cnt++;    LL tmp = cnt;    f[cnt].a = a; f[cnt].b = b;    f[cnt].s = 0;    f[cnt].l = f[cnt].r = NULL;    if(a < b)    {        LL mid = (a + b) / 2;        f[tmp].l = build(a, mid);        f[tmp].r = build(mid + 1, b);    }    return f + tmp;}void insert(node *t, LL a, LL c, LL p){    if(a <= t->a && t->a == t->b) { if(c >= t->s) { t->s = c; t->pos = p; } }    else    {        LL mid = (t->a + t->b) / 2;        if(a <= mid) insert(t->l, a, c, p);        else insert(t->r, a, c, p);        if(t->l->s > t->r->s) t->pos = t->l->pos;        else t->pos = t->r->pos;        t->s = max(t->l->s, t->r->s);    }}LL findMax(node *t, LL a, LL b, LL &pos){    if(a <= t->a && t->b <= b) { pos = t->pos; return t->s; }    LL s = 0;    LL mid = (t->a + t->b) / 2;    LL p1;    if(a <= mid)    {        LL p = findMax(t->l, a, b, p1);        if(p > s)        {            s = p;            pos = p1;        }    }    if(b > mid)    {        LL p = findMax(t->r, a, b, p1);        if(p > s)        {            s = p;            pos = p1;        }    }    return s;}LL shuffle(LL f[], LL n){    LL cnt = 0;    for(LL i = 0; i < n; i++)    {        if(i == 0 || f[i] != f[i - 1]) f[cnt++] = f[i];    }    return cnt;}LL find1(LL x, LL f[], LL n){    LL l = 0, r = n - 1;    while(l < r - 1)    {        LL mid = (l + r) / 2;        if(x <= f[mid]) r = mid;        else l = mid;    }    if(x == f[r]) return r;    return l;}LL find(LL x, LL f[], LL n){    LL l = 0, r = n - 1;    while(l < r - 1)    {        LL mid = (l + r) / 2;        if(x <= f[mid]) r = mid;        else l = mid;    }    if(x == f[l]) return l;    return r;}int main(){    //freopen("test.in", "r", stdin);    scanf("%I64d%I64d", &n, &d);    for(LL i = 0; i < n; i++) scanf("%I64d", &a[i]);    for(LL i = 0; i < n; i++) num[i] = a[i];    LL l = n;    num[n] = -1e15 - 7;    n++;    num[n] = 1e15 + 7;    n++;    num[n + 2] = 0;    n++;    sort(num, num + n);    LL nn = shuffle(num, n);    //for(LL i = 0; i < nn; i++) cout <<num[i] <<" ";    //cout << endl;    memset(dp, 0, sizeof dp);    LL ans = 0;    LL kk;    top = build(0, nn + 1);    for(LL i = 0; i < l; i++)    {        LL pos1, pos2;        //cout << a[i] - d << " " << a[i] <<" "<< a[i] + d << endl;        //cout << num[find1(a[i] - d, num, nn)] << " " << num[find(a[i] + d, num, nn)] <<endl <<endl;        LL a1 = findMax(top, 0, find1(a[i] - d, num, nn), pos1);        LL a2 = findMax(top, find(a[i] + d, num, nn), nn, pos2);        dp[i] = 1;        pre[i] = -1;        if(a1 > a2) { dp[i] += a1; pre[i] = pos1; }        else { dp[i] += a2; pre[i] = pos2; }        //cout << i << " "<< a[i] <<" "<< pre[i] << " " <<dp[i] << endl;        if(dp[i] > ans) { kk = i; ans = max(ans, dp[i]); }        insert(top, find(a[i], num, nn), dp[i], i);    }    l = ans;    while(l)    {        q[l + 1] = kk;        kk = pre[kk];        l--;    }    printf("%I64d\n", ans);    for(LL i = 2; i <= ans + 1; i++) printf("%I64d ", q[i] + 1);    return 0;}