问题 J: 最长路径

题目描述

原题面

There is a directed graph \(G\) with \(n\) vertices and \(m\) edges. The vertices are numbered \(1, 2, ..., n\), and for each \(i\:(1 \le i \le m)\), the \(i\)-th directed edge goes from vertex \(x_i\) to vertex \(y_i\).

\(G\:\) does not contain directed cycles.

Find the length of the longest directed path in \(G\). Here, the length of a directed path is the number of edges in it.

输入要求

The first line contains two integers: number of vertices \(n\:(2 \le n \le 10^5)\) and number of edges \(m\:(1 \le m \le 10^5)\). Each of the next \(m\) lines contains two integer \(x_i, y_i\:(1 \le x_i, y_i \le n)\) that describe the edge of the graph.

输出要求

Print the length of the longest directed path in \(G\).

样例

解法

解法和时间复杂度： (1)

此处时间复杂度为该解法的算法部分的时间复杂度，不是严谨的整题的时间复杂度

法1：拓扑排序 + 动态规划 \(O(n)\)
法2：最短路 - SPFA 算法 \(O(mn^2)\)

法1：拓扑排序 + 动态规划

#include <bits/stdc++.h>
#define int long long int
using namespace std;
const int MAXn = 2e5 + 9;
vector<int> g[MAXn];
int rd[MAXn], dp[MAXn];
queue<int> q;

signed main() {
    int n, m;
    scanf("%lld%lld", &n, &m);
    for(int i = 1, u, v; i <= m; i ++) {
        scanf("%lld%lld", &u, &v);
        u ++; v ++;
        g[u].push_back(v);
        rd[v] ++;
    }
    for(int i = 1; i <= n; i ++)
        if(!rd[i]) q.push(i);
    int ans = 1;
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        for(int next : g[now]) {
            rd[next] --;
            if(!rd[next]) q.push(next);
            dp[next] = max(dp[next], dp[now] + 1);
            ans = max(ans, dp[next]);
        }
    }
    printf("%lld", ans);
    return 0;
}

法2：最短路 - SPFA 算法

读入每条边的值，并记为其负值。使用 SPFA 等可以在负权图中寻找最短路的算法找到图中最短路，并输出路径长度的负值。

由于 SPFA 等算法只能寻找单源最短路，因此需要再从每个点出发，跑一遍 SPFA 等算法，时间复杂度较高，不能通过。

以下提供类似题目 P1807 最长路，只需要求出某两点之间的路径中最长的路径，使用 SPFA 算法，按照上述思路的实现。（该题亦可使用拓扑排序 + 动态规划的方式实现）

#include <bits/stdc++.h>
#define int long long int
using namespace std;
typedef pair<int, int> pii;
const int MAXn = 1000 + 9;
vector<pii> g[MAXn];
int dis[MAXn];
bool ved[MAXn];
queue<int> q;

signed main() {
    for(int i = 0; i <= MAXn - 3; i ++) dis[i] = LLONG_MAX;
    int n, m;
    scanf("%lld%lld", &n, &m);
    for(int i = 1, u, v, w; i <= m; i ++) {
        scanf("%lld%lld%lld", &u, &v, &w);
        g[u].push_back({v, -w});
    }
    dis[1] = 0;
    q.push(1);
    ved[1] = true;
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        ved[now] = false;
        for(pii x : g[now]) {
            int nxt = x.first;
            if (dis[nxt] > dis[now] + x.second) {
                dis[nxt] = dis[now] + x.second;
                if (!ved[nxt]) {
                    q.push(nxt);
                    ved[nxt] = true;
                }
            }
        }

    }
    dis[n] == LLONG_MAX ? printf("-1") : printf("%lld", -dis[n]);
    return 0;
}