洛谷2747（不相交路线、dp）

要点

• 反思：以前是在紫书上做过的……
• $dp[i][j]$是从1引两条路到达i、j的最大值
• 为了不相交，则$dp[i][i]$都是非法的，不转移它，也不用它转移

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#include <string>
#include <vector>
#include <queue>
#include <map>
using namespace std;

const int maxn = 105, inf = 0x3f3f3f3f;
int n, m, ans = 1;
int adj[maxn][maxn], dp[maxn][maxn];
map<string, int> mp;

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		string s;
		cin >> s;
		mp[s] = i;
	}
	for (int i = 1; i <= m; i++) {//graph
		string s, t;
		cin >> s >> t;
		int u = mp[s], v = mp[t];
		adj[u][v] = adj[v][u] = 1;
	}

	dp[1][1] = 1;
	for (int i = 1; i < n; i++)
		for (int j = i + 1; j <= n; j++)//i < j: not intersect
			for (int k = 1; k < j; k++)
				if (adj[k][j] && dp[i][k])
					dp[i][j] = dp[j][i] = max(dp[i][j], dp[i][k] + 1);
				
	for (int i = 1; i < n; i++)
		if (adj[i][n])
			ans = max(ans, dp[i][n]);
	return !printf("%d\n", ans);
}