Intermediate operations can output results but cannot be connected to other operations这句话为什么是错的
时间: 2024-03-17 19:45:22 浏览: 25
这句话是错误的。在Java 8 中,流(Stream)的中间操作(intermediate operations)可以连接多个操作,也可以产生输出结果并将其传递给下一个操作。例如,以下代码中的filter()和map()就是中间操作,并且它们被连接在一起。
```
List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5);
List<Integer> result = numbers.stream()
.filter(n -> n % 2 == 0)
.map(n -> n * 2)
.collect(Collectors.toList());
```
在上面的代码中,filter()和map()都是中间操作,但它们被连接在一起以产生一个最终的结果。因此,中间操作不仅可以输出结果,也可以连接到其他操作。
相关问题
用c++解决Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs). Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem - not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections. You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied. Input The first line contains two integer: N - the number of hubs in the network (2 ≤ N ≤ 1000) and M — the number of possible hub connections (1 ≤ M ≤ 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs. Output Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.
这是一道最小生成树问题,可以使用Prim或者Kruskal算法来解决。下面是使用Kruskal算法的C++代码实现:
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 5;
const int MAXM = 15000 + 5;
struct Edge {
int u, v, w;
Edge() {}
Edge(int u, int v, int w): u(u), v(v), w(w) {}
bool operator<(const Edge& e) const {
return w < e.w;
}
};
int n, m;
int parent[MAXN];
Edge edges[MAXM], mst[MAXM];
int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
void merge(int x, int y) {
parent[find(x)] = find(y);
}
void kruskal() {
for (int i = 1; i <= n; i++) {
parent[i] = i;
}
sort(edges, edges + m);
int cnt = 0;
for (int i = 0; i < m; i++) {
int u = edges[i].u, v = edges[i].v, w = edges[i].w;
if (find(u) != find(v)) {
merge(u, v);
mst[cnt++] = edges[i];
}
if (cnt == n - 1) {
break;
}
}
int max_len = 0;
for (int i = 0; i < cnt; i++) {
max_len = max(max_len, mst[i].w);
}
cout << max_len << endl;
cout << cnt << endl;
for (int i = 0; i < cnt; i++) {
cout << mst[i].u << " " << mst[i].v << endl;
}
}
int main() {
cin >> n >> m;
for (int i = 0; i < m; i++) {
int u, v, w;
cin >> u >> v >> w;
edges[i] = Edge(u, v, w);
}
kruskal();
return 0;
}
```
输入样例:
```
4 5
1 2 1
1 3 2
3 4 3
2 4 4
1 4 5
```
输出样例:
```
4
3
1 2
1 3
3 4
```
intermediate features是什么
Intermediate features是指在计算机视觉领域中,对于图像特征提取的一种方法,它是在低级特征(如像素)和高级特征(如语义概念)之间的中间层次上提取的特征。Intermediate features可以通过卷积神经网络(CNN)等深度学习模型来提取,它们可以表示图像中的局部或全局结构信息,例如边缘、纹理、形状等。Intermediate features在图像分类、目标检测和图像分割等任务中具有广泛的应用。
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