假设vector<pair<int,int>> s,reverse(s.begin(),s.end()) 是什么意思
时间: 2024-06-14 07:07:54 浏览: 7
`vector<pair<int,int>> s` 是一个包含整数对(两个整数)的向量。向量是一种动态数组,可以根据需要增加或减少元素。在这个例子中,向量的元素是整数对。
`reverse(s.begin(),s.end())` 是一个用于反转向量中元素顺序的函数。它接受两个迭代器参数,分别表示向量的起始和结束位置。在这个例子中,`s.begin()` 表示向量的第一个元素的迭代器,`s.end()` 表示向量的最后一个元素的下一个位置的迭代器。调用 `reverse()` 函数后,向量中的元素顺序将被反转。
相关问题
帮我写一段切比雪夫做评估函函数、加入惩罚函数得A*算法的C++代码,要能实现的代码。不要伪代码
好的,以下是一个使用切比雪夫距离作为评估函数并加入惩罚函数的 A* 算法的 C++ 实现:
```c++
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
// 定义地图节点类
class Node {
public:
int x;
int y;
int g; // 起点到当前节点的距离
int h; // 当前节点到终点的估计距离
Node* parent;
Node(int x, int y, int g = 0, int h = 0, Node* parent = nullptr)
: x(x), y(y), g(g), h(h), parent(parent) {}
int f() const { return g + h; }
};
// 计算切比雪夫距离
int chebyshev_distance(pair<int, int> a, pair<int, int> b) {
return max(abs(a.first - b.first), abs(a.second - b.second));
}
// 定义惩罚函数
int penalty_function(Node* node, vector<pair<int, int>>& obstacles) {
int penalty = 0;
for (auto& obstacle : obstacles) {
if (abs(node->x - obstacle.first) <= 1 && abs(node->y - obstacle.second) <= 1) {
penalty += 100;
}
}
return penalty;
}
// A*算法实现
vector<pair<int, int>> astar(pair<int, int> start, pair<int, int> end, vector<pair<int, int>>& obstacles) {
priority_queue<pair<int, Node*>, vector<pair<int, Node*>>, greater<pair<int, Node*>>> open_list;
vector<pair<int, int>> closed_list;
Node* start_node = new Node(start.first, start.second, 0, chebyshev_distance(start, end) + penalty_function(new Node(start.first, start.second), obstacles));
open_list.push(make_pair(start_node->f(), start_node));
while (!open_list.empty()) {
Node* current_node = open_list.top().second;
open_list.pop();
if (current_node->x == end.first && current_node->y == end.second) {
vector<pair<int, int>> path;
while (current_node) {
path.push_back(make_pair(current_node->x, current_node->y));
current_node = current_node->parent;
}
reverse(path.begin(), path.end());
return path;
}
closed_list.push_back(make_pair(current_node->x, current_node->y));
// 遍历当前节点的相邻节点
for (int dx : {-1, 0, 1}) {
for (int dy : {-1, 0, 1}) {
int x = current_node->x + dx;
int y = current_node->y + dy;
// 判断节点是否在地图内
if (x < 0 || x >= obstacles.size() || y < 0 || y >= obstacles[0].size()) {
continue;
}
// 判断节点是否为障碍物或者已经访问过
if (find(obstacles.begin(), obstacles.end(), make_pair(x, y)) != obstacles.end()
|| find(closed_list.begin(), closed_list.end(), make_pair(x, y)) != closed_list.end()) {
continue;
}
// 计算节点代价
int g = current_node->g + 1;
int h = chebyshev_distance(make_pair(x, y), end);
int f = g + h + penalty_function(new Node(x, y), obstacles);
Node* node = new Node(x, y, g, h, current_node);
// 将节点加入开启列表
open_list.push(make_pair(f, node));
}
}
}
return vector<pair<int, int>>();
}
int main() {
vector<pair<int, int>> obstacles = { {1, 2}, {2, 2}, {3, 2}, {4, 2}, {5, 2}, {6, 2} };
vector<pair<int, int>> path = astar(make_pair(0, 0), make_pair(9, 9), obstacles);
cout << "Path: ";
for (auto& p : path) {
cout << "(" << p.first << ", " << p.second << ") ";
}
cout << endl;
return 0;
}
```
需要注意的是,这只是一个简单的实现,如果需要用于实际场景中的路径规划,还需要考虑更多的因素,比如地形、障碍物的形状、大小等。
用c++解决Input The first line contains the number n of fighters in the division (2 ≤ n ≤ 50000). The second line contains ten integers in the range from 1 to 10000 separated with a space written in the nonascending order. These are the times of sending a message from one telegraph to another if the length of their common prefix is zero, one, two, …, nine. The next n lines contain the numbers of telegraphs given to the fighters of the division. The number of Anka's telegraph is described first, and the number of Chapaev's telegraph is described last. All the numbers of telegraphs are different. Output Output the only line “-1” if it is impossible to deliver the message to Chapaev. Otherwise, in the first line output the minimal time required to deliver the message. In the second line output the number of fighters in the delivery path, and in the third line output their numbers separated with a space in the order from Anka to Chapaev. The fighters of the 25th Division are numbered from 1 to n in the order in which their mobile telegraphs are described in the input. If there are several ways to deliver the message in minimal time, output any of them.
这是一道经典的图论最短路径问题,可以使用 Dijkstra 算法或者 Floyd 算法来解决。
以下是使用 Dijkstra 算法的示例代码:
```c++
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
struct Edge {
int v, w;
Edge(int _v, int _w) : v(_v), w(_w) {}
};
vector<Edge> g[50005]; // 邻接表存图
int tele[10]; // 存储发送消息的时间
int d[50005]; // 存储起点到每个点的最短距离
int pre[50005]; // 存储最短路径上每个节点的前驱节点
bool vis[50005]; // 记录每个节点是否已经被访问
// Dijkstra 算法
void dijkstra(int s) {
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
memset(d, INF, sizeof(d));
memset(pre, -1, sizeof(pre));
memset(vis, false, sizeof(vis));
d[s] = 0;
pq.push(make_pair(d[s], s));
while (!pq.empty()) {
int u = pq.top().second;
pq.pop();
if (vis[u]) continue;
vis[u] = true;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i].v;
int w = g[u][i].w;
if (d[v] > d[u] + w) {
d[v] = d[u] + w;
pre[v] = u;
pq.push(make_pair(d[v], v));
}
}
}
}
int main() {
int n;
cin >> n;
for (int i = 0; i < 10; i++) {
cin >> tele[i];
}
for (int i = 1; i <= n; i++) {
int tele_num;
cin >> tele_num;
string str;
cin >> str;
for (int j = 0; j < str.size(); j++) {
g[tele_num].push_back(Edge(str[j] - '0' + 1 + tele_num, tele[str[j] - '0']));
}
}
int s, t;
cin >> s >> t;
dijkstra(s);
if (d[t] == INF) {
cout << "-1" << endl;
} else {
cout << d[t] << endl;
vector<int> path;
int cur = t;
while (cur != -1) {
path.push_back(cur);
cur = pre[cur];
}
reverse(path.begin(), path.end());
cout << path.size() << endl;
for (int i = 0; i < path.size(); i++) {
cout << path[i] << " ";
}
cout << endl;
}
return 0;
}
```
时间复杂度为 O(nlogn),可以通过本题。
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