之前收藏了极客时间的算法训练营3期 共21课，计划每一课写博客来记录学习，主要形式为

方法类型1

题1

题解

题2

题解

方法类型2

题1

题解

……

题目大体来自leetcode 和 acwing

主要记录和理解代码，所以基本完全搬运了视频题解代码，

个人学习感受体现在大致思路的总结和注释上。

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第一题

743. 网络延迟时间

Bellmen -ford

最多n - 1轮，可以处理有负数边的情况

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<int> dist(n + 1, 1e9);
        dist[k] = 0;
        for (int round = 1; round <= n - 1; round++) {
            for (vector<int> time : times) {
                int x = time[0];
                int y = time[1];
                int z = time[2];
                if (dist[x] + z < dist[y])
                    dist[y] = dist[x] + z;
            }
        }
        int ans = 0;
        for (int i = 1; i <= n; i++) ans = max(ans, dist[i]);
        return ans == 1e9 ? -1 : ans;
    }
};

Dijkstra

正边权，O(n2)

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<int> dist(n + 1, 1e9);
        vector<vector<int>> to = vector<vector<int>>(n + 1, vector<int>());
        vector<vector<int>> edge = vector<vector<int>>(n + 1, vector<int>());
        for (vector<int> time : times) {
            int x = time[0];
            int y = time[1];
            int z = time[2];
            to[x].push_back(y);
            edge[x].push_back(z);
        }
        vector<bool> expand(n + 1, false);
        dist[k] = 0;
        for (int round = 0; round < n; round++) {
            int tmp = 1e9;
            int x;
            for (int i = 1; i <= n; i++) {
                if (!expand[i] && dist[i] < tmp){
                    tmp = dist[i];
                    x = i;
                }
            }
            expand[x] = true;
            for (int i = 0; i < to[x].size(); i++) {
                int y = to[x][i];
                int z = edge[x][i];
                if (dist[x] + z < dist[y]){
                    dist[y] = dist[x] + z;
                }
            }
        }
        int ans = 0;
        for (int i = 1; i <= n; i++) ans = max(ans, dist[i]);
        return ans == 1e9 ? -1 : ans;
    }
};

第二题

1334. 阈值距离内邻居最少的城市

floid 得到所有最短距离，再统计。

class Solution {
public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        vector<vector<int>> dist(n, vector<int>(n, 1e9));
        for (int i = 0; i < n; i++) dist[i][i] = 0;
        for (vector<int> edge : edges) {
            int x = edge[0];
            int y = edge[1];
            int z = edge[2];
            dist[x][y] = dist[y][x] = z;
        }
        for (int k = 0; k < n; k++)
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
        int Minneighbour = 1e9;
        int ans;
        for (int i = 0; i < n; i++) {
            int neighbour = 0;
            for (int j = 0; j < n; j++) {
                if (i != j && dist[i][j] <= distanceThreshold){
                    neighbour++;
                }
            }
            if (neighbour <= Minneighbour) {
                Minneighbour = neighbour;
                ans = i;
            }
        }
        return ans;
    }
};

第三题

​​​​​​1584. 连接所有点的最小费用

最朴素的Kruskal,,超时。

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        fa = vector<int>(n);
        for (int i = 0; i < n; i++) fa[i] = i;
        vector<vector<int>> edges;
        for (int i = 0; i < n; i++) 
            for (int j = i + 1; j < n; j++) {
                edges.push_back({i, j, abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])});
            }
        sort(edges.begin(), edges.end(), [](vector<int> A, vector<int> B){
            return A[2] < B[2];
        });
        int ans = 0;
        for (vector<int> edge : edges) {
            int x = edge[0];
            int y = edge[1];
            int z = edge[2];
            x = find(x);
            y = find(y);
            if (x != y) {
                ans += z;
                fa[x] = y;
            }
        }
        return ans;
    }
private:
    vector<int> fa;
    int find(int x) {
        if (fa[x] == x) return x;
        return fa[x] = find(fa[x]);
    }
};