随笔- 23  文章- 0  评论- 0 

补得甚是艰辛

-如非必要,不用STL-患者红小豆上线

参考于:https://www.cnblogs.com/Dillonh/p/11240083.html(为了两个continue还在评论区叨扰了一下,多次试验之后似乎有点明白,趁没被注意偷偷删掉:P)

    https://www.cnblogs.com/tian-luo/p/11254151.html(%%%)

 

 

HDU-6592 Beauty of  unimodal sequence

  调完了诸如中括号里打了另一个数组的下标之类的bug之后,最大的wa点是dp数组大小少打个0

  const xx maxn 一个错全都错,还是每个数组打一次大小 一个错错一个,这是一个问题_(:з」∠)_

  输出格式好要命啊啊啊wa之后还来了发PE

  正文:首先对序列正着跑一边LIS,再反着跑一遍,分别记录下标(ppo-positive,opo-over:P)顺便给po初始化一下;

     原blog开了两个vector,小可寻思着一个数组(叫v吧)加个下标变量就可以复用。

     task1  字典序最小下标序列--开始找峰,字典序最小,所以一旦找到,就用这个。接着开始从峰向前找尽量靠前的同LIS下标的数,看了原blog第一个continue的条件,感觉不会有那种情况,然后就得寸进尺地(x)想去掉第二个continue,然而第二个是有去重作用的_(:з」∠)_查了好久的输出文件突然察觉到这个问题。前面查完了再查后面,后面的部分就是只要符合条件就拿,这样下标最小。stack和v倒腾一下输出就完事了。

     task 2 字典序最大下标序列--开始找峰,找到尽量靠后的那个峰,顺手重新初始化po。往前找,找到就拿,这样下标最大。往后找,(一个迷之懒得用reverse的红小豆就用v模拟了一个stack)进行和task1里的前半部分类似的操作。输出完事。

  细节见码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<stack>
using namespace std;
typedef long long LL;
int n;
int num[300005], ppo[300005], opo[300005], v[300005], d[300005], po[300005];
stack<int>sa;

int main()
{
    while (~scanf("%d", &n)) {

        memset(d, 0x3f, sizeof d);
        d[0] = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &num[i]);
            ppo[i] = lower_bound(d + 1, d + 1 + n, num[i]) - d;
            d[ppo[i]] = num[i];
            po[i] = 0x3f3f3f3f;
        }
        memset(d, 0x3f, sizeof d);
        d[0] = 0;
        for (int i = n; i >= 1; i--) {
            opo[i] = lower_bound(d + 1, d + 1 + n, num[i]) - d;
            d[opo[i]] = num[i];
        }


        int co = 0;
        int now = 1, m = ppo[1] + opo[1];
        for (int i = 2; i <= n; i++)
            if (ppo[i] + opo[i] > m) { now = i; m = opo[i] + ppo[i]; }
        po[ppo[now]] = num[now];
        for (int i = now - 1; i >= 1; i--) {
            if (po[ppo[i] + 1] <= num[i]) continue;
            while (!sa.empty() && ppo[i] >= ppo[sa.top()])po[ppo[sa.top()]] = 0x3f3f3f3f, sa.pop();
            sa.push(i);
            po[ppo[i]] = num[i];
        }
        while (!sa.empty()) v[co++] = sa.top(), sa.pop();
        v[co++] = now;
        for (int i = now + 1; i <= n; i++)
            if (opo[i] == opo[v[co - 1]] - 1 && num[i] < num[v[co - 1]])v[co++] = i;
        for (int i = 0; i < co; i++)printf("%d%c", v[i], " \n"[i == co - 1]);


        co = 0;
        now = 1; m = ppo[1] + opo[1]; po[1] = 0x3f3f3f3f;
        for (int i = 2; i <= n; i++) {
            if (ppo[i] + opo[i] >= m) { now = i; m = opo[i] + ppo[i]; }
            po[i] = 0x3f3f3f3f;
        }
        sa.push(now);
        for (int i = now - 1; i >= 1; i--)
            if (ppo[i] == ppo[sa.top()] - 1 && num[i] < num[sa.top()])sa.push(i);
        while (!sa.empty())v[co++] = sa.top(), sa.pop();
        po[opo[now]] = num[now];
        for (int i = now + 1; i <= n; i++) {
            if (num[i] >= po[opo[i] + 1]) continue;
            while (co && opo[i] >= opo[v[co - 1]])po[opo[v[co - 1]]] = 0x3f3f3f3f, co--;
            v[co++] = i;
            po[opo[i]] = num[i];
        }
        for (int i = 0; i < co; i++)printf("%d%c", v[i], " \n"[i == co - 1]);
    }

    return 0;
}
Beauty of unimodal sequence

 

  (新建之前想着还有什么要写来着但是忘了。。想起来再说)

 

upd: 

HDU-6599 I Love Palindrome String

  今天,你学回文树了吗?https://blog.csdn.net/u013368721/article/details/42100363

  整个学习过程充满指针乱跳,终于是大概知道怎么用了,配上Thor的马拉车板子,开始乱搞。

  大佬题解中非常神奇的部分是在回文树中自加的那个函数,将原回文串中的下标映射回马拉车得到的回文串的长度数组的下标,检查半个回文串的长度,判断是否符合条件

  一开始各种T,学大佬加了一些inline和register之后又wa来wa去,检查了马拉车后来加的两个数组发现开小了_(:з」∠)_

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long LL;

const int MAXN = 300005;
const int N = 26;
int ans[MAXN], rlen[MAXN * 2];
char a[MAXN];
char  str[MAXN * 2];
int R[MAXN*2];

struct Palindromic_Tree {
    int next[MAXN][N], fail[MAXN], cnt[MAXN], len[MAXN], S[MAXN], n, p, last;
    int l[MAXN], r[MAXN];

    inline int newnode(int l) {
        for (register int i = 0; i < N; ++i) next[p][i] = 0;
        cnt[p] = 0;
        len[p] = l;
        return p++;
    }

    void init() {
        p = 0;
        newnode(0);
        newnode(-1);
        last = 0;
        n = 0;
        S[n] = -1;
        fail[0] = 1;
    }

    inline int get_fail(int x) {
        while (S[n - len[x] - 1] != S[n]) x = fail[x];
        return x;
    }

    inline void add(int c) {
        c -= 'a';
        S[++n] = c;
        int cur = get_fail(last);

        if (!next[cur][c]) {
            int now = newnode(len[cur] + 2);
            fail[now] = next[get_fail(fail[cur])][c];
            next[cur][c] = now;
            l[now] = n - len[cur] - 1;
            r[now] = n;

        }
        last = next[cur][c];
        cnt[last] ++;
    }

    void count() {
        for (register int i = p - 1; i >= 0; --i) cnt[fail[i]] += cnt[i];
    }

    void map()
    {
        count();
        for (int i = 2; i < p; i++) {

            int ml = l[i] << 1, mmid = (l[i] + r[i]) >> 1 << 1; 
            int mid = (ml + mmid) >> 1;
            if (mid - rlen[mid] + 1 <= ml)ans[len[i]] += cnt[i];
        }
    }

}pt;


void ma(int m)
{
    memset(R, 0, sizeof(R));
    int n = 0;
    str[n++] = '!';
    str[n++] = '#';
    for (int i = 1; i <= m; i++)
        str[n++] = a[i], str[n++] = '#';
    str[n++] = '#';
    str[n++] = '?';
    int p = 0, mx = 0, ans = 0;
    for (int i = 1; i < n; i++)
    {
        R[i] = mx > i ? min(R[2 * p - i], mx - i) : 1;
        while (str[i + R[i]] == str[i - R[i]]) R[i]++;
        if (R[i] + i > mx)
            mx = i + R[p = i];
        rlen[i] = R[i] - 1;
    }
}


int main()
{

    while (~scanf("%s", a + 1)) {
        pt.init();
        int l = strlen(a + 1);
        for (int i = 1; i <= l; i++)pt.add(a[i]), ans[i] = 0;
        ma(l);
        pt.map();
        for (int i = 1; i <= l; i++)printf("%d%c", ans[i], " \n"[i == l]);

    }

    return 0;
}
I Love Palindrome Strings

  

 

 posted on 2019-08-04 17:28  Nonad  阅读(...)  评论(... 编辑 收藏