Question

为什么我在执行代码时遇到此错误？在其中我试图为图形实现邻接列表。

#include <iostream>
#include<vector>

#include<stdio.h>

using namespace std;

int main()
{
    int t;

    cin >> t;
    while(t--)
    {
        int n,k,i,m=0;
        cin >> n;
        vector<int> mv[n];
        for(i=0;i<n;i++)
        {
            for(m=0;m<n;m++)
            {
                scanf("%d",&k);
                if(k>0)
                    mv[i].push_back(k);
            }
        }
        cout << mv[0].at(0) << " ";
        cout << mv[0].at(1) << " ";
        cout << mv[0].at(2) << " ";
        cout << mv[1].at(1) << " ";
        cout << mv[2].at(1) << " ";
        cout << mv[3].at(0) << " ";
    }
    return 0;
}

Answer 1

您可以在不确定自己处于向量范围内的情况下迭代元素...

替换

cout << mv[0].at(0) << " ";
cout << mv[0].at(1) << " ";
cout << mv[0].at(2) << " ";
cout << mv[1].at(1) << " ";
cout << mv[2].at(1) << " ";
cout << mv[3].at(0) << " ";

人：

for(i=0;i<n;i++)
{
    for ( size_t pos = 0; pos != mv[i].size(); ++pos )
        cout << mv[i].at(pos) << " ";
}

它应该比......更好......

在抛出'std :: out_of_range'的实例后调用终止what（）：vector :: _ M_range_check

1 个答案: