E

E - Petya and Exam

 CodeForces - 832B 

这个题目其实可以不用字典树写,但是因为之前写过poj的一个题目,意思和这个差不多,所以就用字典树写了一遍。

代码还是很好理解的,主要就是哪个findx函数,这个要好好理解。

如果碰到*或着?就要重新标记一下,其他都是一样的,对于?可以跳过好字母,对于*可以跳过若干个不好的字母,

这个就在直接跳过之前加一个判断条件就可以了。

贴一下代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int maxn = 1e5 + 10;
typedef long long ll;
int tree[maxn][30], tot = 0, ans, isgood[1000];
bool flag[maxn];
char good[30], s[maxn], str[maxn];
int add(char *s) {
    int root = 0, id, len = strlen(s);
    for (int i = 0; i < len; i++) {
        if (s[i] == '*') id = 26;
        else if (s[i] == '?') id = 27;
        else id = s[i] - 'a';
        if (!tree[root][id]) tree[root][id] = ++tot;
        root = tree[root][id];
        // printf("root=%d s=%c
", root, s[i]);
    }
    flag[root] = true;
    return root;
}

void findx(char *s, int root, int pos) {
    // printf("%s root=%d pos=%d
", s, root, pos);
    int len = strlen(s);
    if (pos > len) return;
    if (len == pos && flag[root]) {
        ans = 1;
        return;
    }
    if (tree[root][26]) {
        int flag = 1;
        int len1 = strlen(str) - 1;
        int ends = len1 - pos;
        // printf("ends=%d len-ends=%d pos=%d
", ends, len - ends-1, pos);
        for (int i = pos; i <= len - ends - 1; i++) {
            int id = s[i] - 'a';
            if (isgood[id]) {
                flag = 0;
                break;
            }
        }
        // printf("flag=%d pos=%d len-ends=%d
", flag, pos, len - ends);
        if (flag&&len - ends >= pos) findx(s, tree[root][26], len - ends);
        // for (int j = pos; j <= len; j++) {
        //     int id = s[j] - 'a';
        //     if (isgood[id]) break;
        //     findx(s, tree[root][26], j + 1);
        // }
    }
    if (tree[root][27]) {
        int length = strlen(good);
        for (int i = 0; i < length; i++) {
            if (s[pos] == good[i]) findx(s, tree[root][27], pos + 1);
        }
    }
    int id = s[pos] - 'a';
    if (s[pos] <= 'z'&&s[pos] >= 'a'&&tree[root][id]) findx(s, tree[root][id], pos + 1);
}

int main() {
    scanf("%s%s", good, str);
    int len = strlen(good);
    for (int i = 0; i < len; i++) {
        int id = good[i] - 'a';
        isgood[id] = 1;
    }
    int num = add(str);
    int m;
    scanf("%d", &m);
    while (m--) {
        scanf("%s", s);
        ans = 0;
        findx(s, 0, 0);
        if (ans) printf("YES
");
        else printf("NO
");
    }
    return 0;
}
字典树

Wild Words poj 1816

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int maxn = 2e6 + 10;
typedef long long ll;
int tree[maxn][30], tot = 0, ans, p[maxn];
bool flag[maxn], vis[maxn];
char s[maxn], str[maxn];
int add(char *s) {
    int root = 0, id, len = strlen(s);
    for (int i = 0; i < len; i++) {
        if (s[i] == '*') id = 26;
        else if (s[i] == '?') id = 27;
        else id = s[i] - 'a';
        if (!tree[root][id]) tree[root][id] = ++tot;
        root = tree[root][id];
    }
    flag[root] = true;
    return root;
}

void findx(char *s, int root, int pos) {
    // printf("%s root=%d pos=%d
", s, root, pos);
    int len = strlen(s);
    if (pos > len) return;
    if (len == pos && flag[root]) {
        vis[root] = 1;
    }
    if (tree[root][26]) {
        for (int i = pos; i <= len; i++) findx(s, tree[root][26], i);
    }
    if (tree[root][27])    findx(s, tree[root][27], pos + 1);
    int id = s[pos] - 'a';
    if (s[pos] <= 'z'&&s[pos] >= 'a'&&tree[root][id]) findx(s, tree[root][id], pos + 1);
}
vector<int>res;
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++)
    {
        scanf("%s", str);
        p[i]=add(str);
    }
    while(m--)
    {
        scanf("%s", str);
        findx(str, 0, 0);
        res.clear();
        for(int i=1;i<=n;i++) if (vis[p[i]]) res.push_back(i - 1);
        int len = res.size();
        if (len == 0) printf("Not match
");
        else {
            for (int i = 0; i < len - 1; i++) printf("%d ", res[i]);
            printf("%d
", res[len - 1]);
        }
        for (int i = 1; i <= n; i++) vis[p[i]] = 0;
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/EchoZQN/p/11370497.html