program.exe停止处理大输入

Question

这是我的代码

#include<iostream>
#include<map>
#include<vector>
#include<algorithm>
#include<string>
#include<math.h>
#include<bitset>
#include<cstring>
#include<sstream>
#include<iomanip>
#include<queue>
using namespace std;
int d1[] = { 0,0,1,-1,1,-1 ,1,-1 };
int d2[] = { -1,1,0,0,1 ,-1,-1,1 };
string vec[200 + 5];
int counter = 0;
long long int floodfill(int n1, int n2, char c1, char c2,int ss)
{
    if (n1 < 0 || n1 >= ss || n2 < 0||n2>=vec[n1].length())
        return 0;
    if (vec[n1][n2] != c1) return 0;
    long long int ans = 1;
    vec[n1][n2] = c2;
    for (int i = 0; i < 8; i++)
        ans = ans + floodfill(n1 + d2[i], n2 + d1[i], c1, c2,ss);
    return ans;
}
int floodfill2(int n1, int n2, char c1, char c2,int ss)
{
    if (n1 < 0 || n1 >= ss || n2 < 0 || n2 >= vec[n1].length())
        return 0;
    if (vec[n1][n2] == '0')
    {
        floodfill(n1, n2, '0', c1,ss);
        counter++;
    }
    if (vec[n1][n2] !=c1) return 0;
    int ans = 1;
    vec[n1][n2] = c2;
    for (int i = 0; i < 4; i++)
        ans = ans + floodfill2(n1 + d1[i], n2 + d2[i], c1, c2,ss);
    return ans;
}
int main() {
    int n1;
    int n2;
    int co = 1;
    while (true)
    {
        cin >> n1;
        cin >> n2;
        if (n1 == 0 && n2 == 0)
            break;
        getchar();
        int aa = 0;
        for (int i = 0; i < n1; i++)
        {
            char s[300];
            scanf("%s", s);
            string g = "";
            int a = strlen(s);
            for (int j = 0; j <a; j++)
            {
                string temp = "";
                if (s[j] == '0')
                    temp = "0000";
                else if (s[j] == '1')
                    temp = "0001";
                else if (s[j] == '2')
                    temp = "0010";
                else if (s[j] == '3')
                    temp = "0011";
                else if (s[j] == '4')
                    temp = "0100";
                else if (s[j] == '5')
                    temp = "0101";
                else if (s[j] == '6')
                    temp = "0110";
                else if (s[j] == '7')
                    temp = "0111";
                else if (s[j] == '8')
                    temp = "1000";
                else if (s[j] == '9')
                    temp = "1001";
                else if (s[j] == 'a')
                    temp = "1010";
                else if (s[j] == 'b')
                    temp = "1011";
                else if (s[j] == 'c')
                    temp = "1100";
                else if (s[j] == 'd')
                    temp = "1101";
                else if (s[j] == 'e')
                    temp = "1110";
                else if (s[j] == 'f')
                    temp = "1111";
                g =g+ temp;
            }
            vec[aa] = g;
            aa++;
        }
        floodfill(0, 0, '0', '.',aa);
        string ans = "";
        for (int i = 0; i < aa; i++)
        {
            for (int j = 0; j < vec[0].length(); j++)
            {
                counter = 0;
                if (vec[i][j] == '1')
                {
                    floodfill2(i, j, '1', '.',aa);
                    if (counter == 0)
                        ans = ans + 'W';
                    else if (counter == 1)
                        ans = ans + 'A';
                    else if (counter == 2)
                        ans = ans + 'K';
                    else if (counter == 3)
                        ans = ans + 'J';
                    else if (counter == 4)
                        ans = ans + 'S';
                    else if (counter == 5)
                        ans = ans + 'D';
                }
            }
        }
        sort(ans.begin(), ans.end());
        cout << "Case " << co  << ": " << ans << endl;
        co++;
    }
}

它在这个输入上工作

6 25
0000000000000000000000000
0000000000000000000000000
00001fe0000000000007c0000
00003fe0000000000007c0000
0000000000000000000000000
0000000000000000000000000

但是停止了对这个的工作

200 50
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Answer 1

您有以下代码：

string vec[200 + 5];
...
if (vec[n1][n2] != c1) return 0;

我看不出你在检查这个数组是否永远不会越界，即代码是否试图访问大于205的元素。

我预计，随着你的大量投入，你将失去债券。

一种简单的检查方法是运行调试构建。一个好的调试c运行库应该会给你一个显式的越界错误。

否则，您可以添加以下内容：

#define MAXSIZE 204
string vec[ MAXSIZE];
...
if( n1 >= MAXSIZE )
{
   cout << "ERROR: Input too bign";
   exit(1);
}