Question

我的任务是找到可以作为服务器运行的计算机，因为它具有到其他计算机的最短路径。

输入第一行有多台计算机。其他行有连接的计算机数量。最后一行总是0 0 例如：

当所有计算机获得消息时，输出第一行包含最少时间（跳转）。第二行包含可用作服务器的所有计算机数量。例如：

2
1 4

在我的本地计算机上，正在运行。但是当我在大学服务器中发送我的代码进行测试时，2/5公共测试正在运行，但是3/5给了我错误：

Caught fatal signal 11
stderr  
*** Error in `solution': corrupted size vs. prev_size: 0x0000000001a36940 ***
======= Backtrace: =========
[0x4bb811]
[0x4c47bb]
[0x4c7d87]
[0x42688e]
[0x429f47]
[0x42aa4d]
[0x400817]
[0x49a0b6]
[0x49a2aa]
[0x401039]
======= Memory map: ========
00400000-005a3000 r-xp 00000000 fd:01 776099                             /box/solution
007a2000-007ab000 rw-p 001a2000 fd:01 776099                             /box/solution
007ab000-007b0000 rw-p 00000000 00:00 0 
01a1f000-01a42000 rw-p 00000000 00:00 0                                  [heap]
2b1ae7340000-2b1ae7349000 rw-p 00000000 00:00 0 
7fffd3bb6000-7fffd3bd7000 rw-p 00000000 00:00 0                          [stack]
7fffd3bfb000-7fffd3bfe000 r--p 00000000 00:00 0                          [vvar]
7fffd3bfe000-7fffd3c00000 r-xp 00000000 00:00 0                          [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0                  [vsyscall]

我找不到自己的错误，希望有人可以提供帮助。我的代码：

#include <fstream>
using namespace std;

int BreathFirstSearch(int** matrix, int size, int row){//function to walk around matrix and find every path from vertex to vertex
    int pathLen=0;
    int q[size+1], dis[size+1], queue_input = 0, queue_output = 0, look_at;
    bool stop = false;
    for(int i=1;i<=size+1;i++){
        q[i]=0;//queue for element I need to look at
        dis[i]=-1; //to now if I have looked at it and to count how many jumps there are between row vertex and every other vertex
    }
    dis[row] = 0;//the row from matrix can go to itself with 0 jumps
    q[queue_input]=row;
    while(q[queue_output]!=0){
        look_at=q[queue_output];
        queue_output++;
        if(!stop){
            for(int col=1; col<=size; col++){//walking through every matrix collon
                if(matrix[look_at][col]==1 && dis[col]==-1){//if matrix collon has 1 to show that there is string between it and the row element. also check if it have already been ir queue
                    queue_input++;
                    q[queue_input]=col;//add collon to queue
                    dis[col]=dis[look_at]+1;//add +1 to path from row to collon element
                }
            }
        }
        int s=0;
        for(int i=1; i<=size; i++){
            for(int k=1; k<=size; k++){
                if(q[k]==i){
                    s++;//count how many elements row has
                    break;
                }
            }
        }
        if(s==size) stop=true; //if row has as much elements as matrix size, then I know that there wont be more and can stop
    }
    int maxDis = 0;
    for (int i = 1; i <= size; i++){
        if (dis[i] > maxDis){
            maxDis = dis[i];
        }
    }
    return maxDis;//return only farthest distance
}

int main(){
    ifstream fin ("serveris.in");
    ofstream fout ("serveris.out");
    if (fin.is_open()){
        int size, x, y;
        fin >> size; //read size of matrix from file
        int** matrix = new int*[size];
        for(int i = 1; i <= size; ++i){
            matrix[i] = new int[size];
        }
        int *result = new int[size];
        for(int i=1; i<=size; i++){//fill matrix with zeros
            for(int k=1; k<=size; k++){
                matrix[i][k]=0;
            }
        }

        while(fin.good()){
            fin >> x >> y;
            if(x!=0){//fill 1 in their places
                matrix[x][y]=1;
                matrix[y][x]=1;
            }
        }

        for(int row=1; row<=size; row++){//go through every row of matrix
            result[row]=BreathFirstSearch(matrix, size, row);//in result array I put in only farthest path
        }

        int min = result[1];
        for(int i=1; i<=size; i++){//find shortest path between vertex
            if(result[i]<min)min=result[i];
        }
        fout<<min<<endl;
        for(int i=1; i<=size; i++){//find in witch row/s this path was
            if(result[i]==min)fout<< i<<' ';
        }
    for(int i=1; i<=size; i++) delete[] matrix[i];
        delete[] matrix;
        delete[] result;
    }
    fin.close();
    fout.close();
}

示例是在服务器上给出错误。这个是有效的：

找到顶点之间的最短路径，但在服务器中有错误

0 个答案: