<p>There is a <strong>directed graph</strong> of <code>n</code> colored nodes and <code>m</code> edges. The nodes are numbered from <code>0</code> to <code>n - 1</code>.</p>

<p>You are given a string <code>colors</code> where <code>colors[i]</code> is a lowercase English letter representing the <strong>color</strong> of the <code>i<sup>th</sup></code> node in this graph (<strong>0-indexed</strong>). You are also given a 2D array <code>edges</code> where <code>edges[j] = [a<sub>j</sub>, b<sub>j</sub>]</code> indicates that there is a <strong>directed edge</strong> from node <code>a<sub>j</sub></code> to node <code>b<sub>j</sub></code>.</p>

<p>A valid <strong>path</strong> in the graph is a sequence of nodes <code>x<sub>1</sub> -&gt; x<sub>2</sub> -&gt; x<sub>3</sub> -&gt; ... -&gt; x<sub>k</sub></code> such that there is a directed edge from <code>x<sub>i</sub></code> to <code>x<sub>i+1</sub></code> for every <code>1 &lt;= i &lt; k</code>. The <strong>color value</strong> of the path is the number of nodes that are colored the <strong>most frequently</strong> occurring color along that path.</p>

<p>Return <em>the <strong>largest color value</strong> of any valid path in the given graph, or </em><code>-1</code><em> if the graph contains a cycle</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet1.png" style="width: 400px; height: 182px;" /></p>

<pre>
<strong>Input:</strong> colors = &quot;abaca&quot;, edges = [[0,1],[0,2],[2,3],[3,4]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The path 0 -&gt; 2 -&gt; 3 -&gt; 4 contains 3 nodes that are colored <code>&quot;a&quot; (red in the above image)</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet2.png" style="width: 85px; height: 85px;" /></p>

<pre>
<strong>Input:</strong> colors = &quot;a&quot;, edges = [[0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is a cycle from 0 to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == colors.length</code></li>
	<li><code>m == edges.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= m &lt;= 10<sup>5</sup></code></li>
	<li><code>colors</code> consists of lowercase English letters.</li>
	<li><code>0 &lt;= a<sub>j</sub>, b<sub>j</sub>&nbsp;&lt; n</code></li>
</ul>
class Solution {
public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n = colors.size(), m = edges.size();
        vector<vector<int>> g(n);
        vector<int> d(n); // d[i]表示i点的入度
        for (auto& e: edges) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            d[b] ++;
        }

        // 用拓扑排序来判断图中是否有环
        vector<int> p; // 存储bfs的递推序列
        queue<int> q;
        for (int i = 0; i < n; i ++)
            if (!d[i]) q.push(i); // 入度为0，入队
        while (q.size()) {
            auto t = q.front();
            q.pop();
            p.push_back(t);
            for (int v: g[t]) {
                if (! -- d[v]) q.push(v); // 若删去t点后，v点入度为0，则入队
            }
        }
        if (p.size() < n) return -1; // 有环

        vector<vector<int>> f(n, vector<int>(26)); // f[i][j]表示以i为终点的所有路径中，包含颜色j的数量的最大值
        int res = 0;
        for (int i: p) {
            int c = colors[i] - 'a'; // c为i点的颜色
            f[i][c] = max(f[i][c], 1); // f[i][c]为以i为终点的路径中，包含颜色c的最多数量，至少也有1个
            
            // 枚举26种颜色j
            for (int j = 0; j < 26; j ++) {
                // 枚举i的每一个邻点k
                for (int k: g[i]) {
                    int t = 0;
                    if (colors[k] - 'a' == j) t = 1; // k点的颜色是j，则t = 1
                    // 根据f[i][j]递推f[k][j]
                    f[k][j] = max(f[k][j], f[i][j] + t);
                }
                res = max(res, f[i][j]);
            }
        }
        return res;
    }
};

/*
首先，判断图中是否有环，若有环，直接返回-1；
然后，枚举26种颜色，计算每条路径上包含该颜色的最多数量是多少；（拓扑图的最长路问题）

f[i][j]表示以i为终点的所有路径中，包含颜色j的数量的最大值是多少。
对于所有可以到达k的i，则f[k][j] = max(f[k][j], f[i][j] + t), 若k点的颜色是j，则t = 1， 否则t = 0。
*/
152363