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bzoj4896 [Thu Summer Camp2016]补退选

2024-09-26 13:52:38   215人阅读

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4896

【题解】

隔壁thusc画风怎么这么正常啊。。

三个操作：插入字符串，删除字符串，询问以某字符串为前缀，最早什么时候超过了d个。

用trie插入然后拿个vector维护第一次超过x个的时间。

vector中总的元素为60*10w=600w个。所以可以承受。

然后我们发现非常坑。。

我们只能用指针写省空间。。然后开了600w的trienode还是MLE了。。

改成400w就过了（？）看来没去卡。。

技术分享 
# include <map>
# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 4e6 + 10, N = 310;
const int mod = 1e9+7;

# define RG register
# define ST static

char str[N];

struct trienode {
    trienode *ch[26];
    int w;
    vector<int> v;
}T[M];
int cnt = 1;

int nQ;
inline void add(char *str, int d) {
    trienode *p = &T[1], *q;
    for (int i=0, c; str[i]; ++i) {
        c = str[i] - ‘a‘;
        if(p->ch[c] == 0) {
            ++cnt;
            q = &T[cnt];
            p->ch[c] = q;
        }
        p = p->ch[c];
        p->w += d;
        if((p->v).size() < p->w) p->v.push_back(nQ);
    }
}

inline int query(int d) {
//    printf("d = %d\n", d);
    trienode *p = &T[1], *q;
    for (int i=0, c; str[i]; ++i) {
        c = str[i] - ‘a‘;
        if(p->ch[c] == 0) return -1;
        p = p->ch[c];
//        printf("%d\n", p->v.size());
    }
    if(p->v.size() > d) return p->v[d];
    else return -1;
}


int main() {
    int q, opt, a, b, c;
    cin >> q;
    int ans = 0;
    for (nQ = 1; nQ <= q; ++nQ) {
        scanf("%d", &opt); 
        scanf("%s", str);
        if(opt == 1) add(str, 1);
        else if(opt == 2) add(str, -1);
        else {
            scanf("%d%d%d", &a, &b, &c);
            ans = query(((ll)a*abs(ans)+b)%c);
            printf("%d\n", ans);
        }
    }
    return 0;
}
View Code

后来看了看popoqqq的博客发现有很多ch[..]是没有用的。

那么拿map存有用的和没用的。。每个节点开个map即可。

这样理论多个log还更快？

技术分享 
# include <map>
# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 6e6 + 10, N = 310;
const int mod = 1e9+7;

# define RG register
# define ST static

char str[N];

struct trienode {
    map<int, trienode*> ch;
    int w;
    vector<int> v;
}T[M];
int cnt = 1;

int nQ;
inline void add(char *str, int d) {
    trienode *p = &T[1], *q;
    for (int i=0, c; str[i]; ++i) {
        c = str[i] - ‘a‘;
        if(p->ch[c] == 0) {
            ++cnt;
            q = &T[cnt];
            p->ch[c] = q;
        }
        p = p->ch[c];
        p->w += d;
        if((p->v).size() < p->w) p->v.push_back(nQ);
    }
}

inline int query(int d) {
//    printf("d = %d\n", d);
    trienode *p = &T[1], *q;
    for (int i=0, c; str[i]; ++i) {
        c = str[i] - ‘a‘;
        if(p->ch[c] == 0) return -1;
        p = p->ch[c];
//        printf("%d\n", p->v.size());
    }
    if(p->v.size() > d) return p->v[d];
    else return -1;
}


int main() {
    int q, opt, a, b, c;
    cin >> q;
    int ans = 0;
    for (nQ = 1; nQ <= q; ++nQ) {
        scanf("%d", &opt); 
        scanf("%s", str);
        if(opt == 1) add(str, 1);
        else if(opt == 2) add(str, -1);
        else {
            scanf("%d%d%d", &a, &b, &c);
            ans = query(((ll)a*abs(ans)+b)%c);
            printf("%d\n", ans);
        }
    }
    return 0;
}
View Code

技术分享

（下一个是用map，上一个是直接400w个trienode开26个ch）

 

bzoj4896 [Thu Summer Camp2016]补退选


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