CF919D Substring

Posted 2020-10-25 王宜鸣

思路：

拓扑排序过程中dp。若图有环，返回-1。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 300005;
 4 vector<int> G[MAXN];
 5 int in[MAXN], n, m;
 6 string value;
 7 int dp[MAXN][26];
 8 bool solve()
 9 {
10     queue<int> q;
11     for (int i = 1; i <= n; i++) 
12     {
13         if (!in[i]) 
14         {
15             q.push(i); dp[i][value[i - 1] - ‘a‘] = 1;
16         }
17     }
18     while (!q.empty())
19     {
20         int x = q.front(); q.pop();
21         for (int i = 0; i < G[x].size(); i++)
22         {
23             int to = G[x][i];
24             for (int k = 0; k < 26; k++) 
25             {    
26                 if (k == value[to - 1] - ‘a‘) dp[to][k] = max(dp[to][k], dp[x][k] + 1);
27                 else dp[to][k] = max(dp[to][k], dp[x][k]);
28             }
29             in[to]--;
30             if (!in[to]) q.push(to);
31         }
32     }
33     for (int i = 1; i <= n; i++) if (in[i]) return false;
34     return true;
35 }
36 int main()
37 {
38     int x, y;
39     cin >> n >> m >> value;
40     set<pair<int, int>> s;
41     bool flg = true;
42     for (int i = 0; i < m; i++) 
43     { 
44         cin >> x >> y;
45         if (x == y) flg = false;
46         pair<int, int> p(x, y);
47         s.insert(p);
48     }
49     if (!flg) { cout << -1 << endl; return 0; }
50     for (auto p: s) { G[p.first].push_back(p.second); in[p.second]++; }
51     if (!solve()) cout << -1 << endl;
52     else
53     {
54         int maxn = 0;
55         for (int i = 1; i <= n; i++)
56         {
57             for (int j = 0; j < 26; j++)
58                 maxn = max(maxn, dp[i][j]);
59         }
60         cout << maxn << endl;
61     }
62     return 0;
63 }