报错:streetlamp varies from 3m to 9m, $H2 (3\<
时间: 2024-03-07 22:49:43 浏览: 15
这个错误是因为在LaTeX中小于号 `<` 要用反斜杠 `\textless` 或者命令 `\lt` 来表示,否则会被解释为数学符号。同时,如果你想要在正文中输入数学公式,需要使用数学模式,可以使用 `$...$` 或者 `\(...\)` 将公式包裹起来。
因此,你的代码应该修改为:
```
streetlamp varies from 3m to 9m, \(H2 (3\textless x\textless 9)\).
```
或者
```
streetlamp varies from 3m to 9m, $H2 (3< x< 9)$.
```
这样就可以正确编译了。
相关问题
from origin 'http://localhost:8081' has been blocked by CORS policy
The CORS (Cross-Origin Resource Sharing) policy is a security measure implemented by web browsers to restrict access to resources (such as APIs) on a different origin (domain, protocol, or port) than the one the web page is served from.
In your case, it seems that you are making a request from 'http://localhost:8081' to a different origin, and the server is blocking that request due to CORS policy restrictions. To resolve this issue, you can try the following solutions:
1. Enable CORS on the server-side: If you have control over the server-side code, you can configure the server to include the appropriate CORS headers in the response. These headers allow the browser to determine if it should allow the request. The minimal required headers are 'Access-Control-Allow-Origin' and 'Access-Control-Allow-Methods'. You can set the value of 'Access-Control-Allow-Origin' to '*' to allow requests from any origin during development. However, in production, it's recommended to set it to the specific origin of your application.
2. Use a proxy server: If you don't have control over the server-side code or enabling CORS is not feasible, you can set up a proxy server that acts as an intermediary between your client-side code and the server. Your client-side code will make requests to the proxy server, and the proxy server will forward those requests to the actual server. Since the proxy server will be on the same origin as your client-side code, you won't face any CORS issues.
3. Disable CORS in the browser: Although not recommended for production environments, during development, you can disable CORS checks in your browser. Keep in mind that this is a temporary solution and should not be used in a production environment. The method to disable CORS varies between browsers, so you need to search for instructions specific to your browser.
Remember to consider security implications when implementing any solution and ensure that proper CORS configurations are applied in a production environment.
The following problem is solved by branch and bound method and the C++ code is given Li wants to travel around the country by car. He finds that the price of gasoline varies from city to city during the trip. It is obvious that if he can adopt a reasonable way to fill up, he will save the cost of the whole trip. Assuming the car starts with an empty tank, it will use one unit of gas for every unit of distance it travels. Please help Xiao Li find the most economical way to complete his entire road trip. Input requirements: The first line of the input contains two integers n (1 ≤ n ≤ 1000) and m (0 ≤ m ≤ 10000), which represent the number of cities and the number of roads, respectively. The following line contains n integers, each representing the price of gasoline in n cities. The following m lines, each containing three integers u,v,d, represent the city u, and the distance between v is d (1 ≤ d ≤ 100). The last line contains three integers c (1 ≤ c ≤ 100),s, and e, which represent the capacity of the tank, the city of departure, and the city of last arrival, respectively. Output requirements: If it is possible to help Xiao Li find a route from s to e that is most economical, the corresponding cost will be output; otherwise, "impossible" will be output. Sample input: 5 5 10, 10, 20, 12, 13 0, 1, 9 0, 2, 8 One, two, one 1, 3, 11 Two, three, seven Ten zero three Sample output: 170
以下是使用分支界限法解决该问题的C++代码:
```c++
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 1010, M = 20010;
int n, m, c, s, e;
int p[N], h[N], e[M], ne[M], w[M], idx;
int dist[N][110]; // dist[i][j]表示从起点到i,油量为j时的最短路
bool st[N][110];
struct Node
{
int u, d, c;
bool operator< (const Node& t) const
{
return d > t.d;
}
};
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
priority_queue<Node> heap;
heap.push({s, 0, 0});
dist[s][0] = 0;
while (heap.size())
{
auto t = heap.top();
heap.pop();
int u = t.u, d = t.d, c = t.c;
if (st[u][c]) continue;
st[u][c] = true;
for (int i = h[u]; ~i; i = ne[i])
{
int j = e[i], v = w[i];
if (c + j <= c && dist[j][c + j] > dist[u][c] + p[j] * (c + j - d))
{
dist[j][c + j] = dist[u][c] + p[j] * (c + j - d);
heap.push({j, dist[j][c + j], c + j});
}
}
}
int res = 0x3f3f3f3f;
for (int i = 0; i <= c; i ++ ) res = min(res, dist[e][i]);
if (res == 0x3f3f3f3f) return -1;
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i ++ ) scanf("%d", &p[i]);
memset(h, -1, sizeof h);
while (m -- )
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c), add(b, a, c);
}
scanf("%d%d%d", &c, &s, &e);
int res = dijkstra();
if (res == -1) puts("impossible");
else printf("%d\n", res);
return 0;
}
```
思路解析:
- 由于题目要求求最优解,因此我们可以想到使用单源最短路径算法来求解,这里我们采用Dijkstra算法。
- 由于油量对于最短路的计算有影响,因此我们可以使用二维数组`dist[i][j]`来表示从起点到i,油量为j时的最短路。
- 在Dijkstra算法中,每次从堆中取出距离起点最近的节点`u`,在遍历`u`的邻居节点`v`时,我们需要考虑两种情况:(1)当前油量`c`可以到达节点`v`,则直接更新`dist[v][c]`;(2)当前油量`c`无法到达节点`v`,则需要选择一个新的油量`c'`来到达节点`v`,那么节点`v`的油量将变为`c' - dis(u, v)`,因此我们可以将更新`dist[v][c']`的过程转化为更新`dist[v][c]`的过程,即`dist[v][c'] = dist[u][c] + p[v] * (c' - dis(u, v))`,其中`p[v]`表示节点`v`的油价。
- 由于油量是一个连续的值,因此我们需要使用bool数组`st[i][j]`来标记节点`i`的油量是否被访问过,防止重复访问。
- 最后,我们在`dist[e][i]`中取最小值即为答案。
时间复杂度:$O(mc\log n)$
其中,$m$为边数,$c$为油量,$n$为点数。