用ChatGPT把图片转换成文本，方便阅读。明天思考一下来写代码。

1. Optimal Transfer

There are n servers in the network arranged in ascending order of their capacity. In the array capacity, the capacity of the ith server is capacity[i], where 0 ≤ i < n.

The distance between two servers, i and j, is defined as the absolute difference in their capacities: |capacity[i] - capacity[j]|. For each server i, the closest server j is the one with the smallest distance to i, and this closest server is unique.

To manage the network, the following operations can be done to server x:

• Connect to any server y at a cost of |capacity[x] - capacity[y]| units.
• Connect to the closest server of x for a fixed cost of one unit.

Given m queries, each defined by two integers fromServer[i] and toServer[i], find the minimum cost required to connect from fromServer[i] to toServer[i] for each query. The connection can be either direct or routed through a server z.

Note:

• The values in the capacity array are distinct.

Example

n = 3, m = 3  
capacity = [2, 7, 10]  
fromServer = [0, 1, 2]  
toServer = [2, 2, 1]  

Optimal connections for each query

• For 0 → 1: server 1 is closest because |2 - 7| < |2 - 10|
• For 1 → 2: server 2 is closest because |7 - 10| < |7 - 2|
• For 2 → 1: server 1 is closest because |10 - 7| < |10 - 2|

下面是我的理解。
这题看起来是在有向图里找最短路径，可以用Dijkastra算法。只不过有一个特殊之处，那就是距离最近的两个节点之间的权重会被固定成1. 我觉得可以在构建有向图的过程中把这个特殊要求算进去，也就是说如果距离是最近的，就把权重设为1. 然后再用Dijkastra算法去求最短路径。
让大语言模型写了一个C++的版本，凑合着看看吧。之后有空我会来写点测试。

#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <limits>
#include <unordered_map>

using namespace std;

const int INF = numeric_limits<int>::max();

// Build the graph with special closest-server rule
vector<vector<pair<int, int>>> buildGraph(const vector<int>& capacity) {
    int n = capacity.size();
    vector<vector<pair<int, int>>> graph(n);

    for (int i = 0; i < n; ++i) {
        int minDiff = INF, closest = -1;

        // Find closest server to i
        for (int j = 0; j < n; ++j) {
            if (i == j) continue;
            int diff = abs(capacity[i] - capacity[j]);
            if (diff < minDiff) {
                minDiff = diff;
                closest = j;
            }
        }

        for (int j = 0; j < n; ++j) {
            if (i == j) continue;
            int cost = abs(capacity[i] - capacity[j]);
            if (j == closest) cost = 1;  // Apply discount
            graph[i].push_back({j, cost});
        }
    }

    return graph;
}

// Dijkstra from one source
int dijkstra(int n, int src, int dst, const vector<vector<pair<int, int>>>& graph) {
    vector<int> dist(n, INF);
    dist[src] = 0;

    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> pq;
    pq.push({0, src});

    while (!pq.empty()) {
        auto [d, u] = pq.top(); pq.pop();
        if (u == dst) return d;

        for (auto [v, cost] : graph[u]) {
            if (dist[u] + cost < dist[v]) {
                dist[v] = dist[u] + cost;
                pq.push({dist[v], v});
            }
        }
    }

    return dist[dst];
}

int main() {
    // Example input
    vector<int> capacity = {2, 7, 10};
    vector<int> fromServer = {0, 1, 2};
    vector<int> toServer = {2, 2, 1};

    int n = capacity.size();
    auto graph = buildGraph(capacity);

    for (int i = 0; i < fromServer.size(); ++i) {
        int cost = dijkstra(n, fromServer[i], toServer[i], graph);
        cout << "Cost from server " << fromServer[i]
             << " to server " << toServer[i] << " = " << cost << endl;
    }

    return 0;
}

请问是要求c++作答吗

可以用python吗

有大佬做了吗
我的第一题和上面差不多 djiakstra
第二题：滑动窗口

int count_substring_vowel(string& s) {
		int res = 0;
		unordered_set<char> vowels = { 'a', 'e', 'i', 'o', 'u' };
		unordered_map<char, int> count;
		int l = 0, r = 0, head = 0;
		int n = s.size();
		while (r < n) {
			auto cur = s[r++];
			if (!vowels.count(cur)) {
				head = l = r;
				count.clear();
			}
			else {
				count[cur]++;
				if (count.size() == 5) {
					while (count.size() == 5) {
						
						auto del = s[l++];
						count[del]--;
						if (count[del] == 0) count.erase(del);
					}
					res += l - head;
					for (int i = head; i < l;i++) {
						cout << string(s.begin() + i, s.begin() + r) << endl;
					}
				}
			}

		}
		return res;
	}

第三题：dp

int findMaxVol(vector<int>& starts, vector<int>& durations, vector<int>& vols) {
		vector<vector<int>> intervals;
		intervals.reserve(starts.size());
		for (int i = 0; i < starts.size(); i++) {
			intervals.push_back({ starts[i] + durations[i], starts[i], vols[i]});
		}
		sort(intervals.begin(), intervals.end());
		vector<int> dp(intervals.size() + 1, 0);
		for (int i = 0; i < intervals.size(); i++) {
			int res = dp[i];
			int prev = bs(intervals, intervals[i][1]);
			//if (prev == -1 || prev >= i) continue;
			res = max(res, intervals[i][2] + dp[prev + 1]);
			dp[i + 1] = res;
		}
		return dp.back();

	}
	int bs(vector<vector<int>>& intervals, int target) {
		int l = 0, r = intervals.size();
		while (l < r) {
			int mid = (l + r) >> 1;
			int mid_num = intervals[mid][0];
			if (mid_num == target) r = mid ;
			if (mid_num > target) r = mid ;
			if (mid_num < target) l = mid + 1;
		}
		return l - 1;
	}

test case不够不知道正不正确 有没人交流下