2019ICPC上海站

https://ac.nowcoder.com/acm/contest/4370

 

B题

判断是不是前缀编码T组样例,N个字符串,判断有没有一个是另一个的前缀

 

Trie

标记这个节点的编号是否出现过两次

注意,一定要有ed,比如说第二组测试样例,如果没有ed,5,59此时5已经走过了

#include <bits/stdc++.h>
using namespace std;
const int maxn = 10005 * 10;
string str;
int t,n;
int flag;
int trie[maxn][26],id = 1,vis[maxn],ed[maxn];
void insert(string s) {
    int p = 1;
    for(int i = 0; i < s.size(); i++) {
        int ch = s[i] - '0';
        if (trie[p][ch]== -1)
            trie[p][ch] = ++id;
        p = trie[p][ch];
        if(ed[p])
            flag = 0;
        vis[p]++;
    }
    if(vis[p] > 1)
        flag = 0;
    ed[p]++;

}
int main(){
    //freopen("in","r",stdin);
    ios::sync_with_stdio(0);
    cin >> t;
    for(int i = 1; i <= t;i++){
        cout << "Case #" << i << ": ";
        cin >> n;
        flag = 1;
        memset(vis,0, sizeof(vis));
        memset(ed,0, sizeof(ed));
        memset(trie,-1, sizeof(trie));
        id = 1;
        for(int j = 1; j <= n; j++) {
            cin >> str;
            insert(str);
        }
        if(flag)
            cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}
View Code

 

 

 

 

K Color Graph

 

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 int t, n, m;
 5 int u[205], v[205];
 6 int vis[20];
 7 int ans;
 8 
 9 void dfs(int x) {
10     if (x == n) {
11         int cnt = 0;
12         for (int i = 1; i <= m; i++)
13             cnt += (vis[u[i]] ^ vis[v[i]]);
14         ans = max(ans, cnt);
15         return;
16     }
17 
18     vis[x] = 1;
19     dfs(x + 1);
20     vis[x] = 0;
21 
22     dfs(x + 1);
23 }
24 
25 int main() {
26     //freopen("in", "r", stdin);
27     ios::sync_with_stdio(0);
28     cin >> t;
29     for (int j = 1; j <= t; j++) {
30         cin >> n >> m;
31         ans = 0;
32         for (int i = 1; i <= m; i++)
33             cin >> u[i] >> v[i];
34         dfs(1);
35         cout << "Case #" << j << ": " << ans << endl;
36     }
37     return 0;
38 }
View Code

 

posted @ 2020-03-13 14:16  Hazelxcf  阅读(638)  评论(0编辑  收藏  举报