海岸长度，kattis

Question

我正在尝试解决我在网站 https://open.kattis.com/problems/coast 上发现的一个问题. Tl;dr 版本的问题是，对于给定的景观 map ，我应该打印出海岸线的长度(没有内岛)。

我收到 0/26 分，但我不知道为什么，我已经测试过，据我检查，它有效。我假设它不编译，但如果是这样，那是为什么呢？它为我编译得很好。

#include <stdio.h>


int edgeCount(int, int, char*);
int topToBottomCount(int, int, char*);
int leftToRightCount(int, int, char*);
int removingInsides(int, int, char*);

int main()
{
    int n = 0; // number of strings
    int m = 0; // strings lenghts
    //printf("Enter N(number of strings) x M(strings lenght): ");
    scanf("%d", &n);
    scanf("%d", &m);

    char coast[1024];
    for(int i = 0; i < n; i++){ 
        scanf("%s", coast+i*m); // adding strings to char coast[1024], making array of ones and zeroes // e.g we are adding 3x4 strings - 111100001111
    }                                                                                                  // it can also be looked as  1111
                                                                                                                            //      0000        - matrix
    int coasts = edgeCount(n, m, coast);                                                                                    //      1111 
    coasts += topToBottomCount(n, m, coast);
    coasts += leftToRightCount(n, m, coast);
    coasts -= removingInsides(n, m, coast);

    printf("%d - coasts\n", coasts);

    return 0;
}

int edgeCount(int n, int m, char *coast){ // if 1 is placed at the edge of the "map", it is 1 coast (2 if it is at corner)
    int edgeCoast = 0;

    for(int i = 0; i < m; i++){     // top edges
        if(coast[i] == '1')
            edgeCoast++;
    }

    for(int i = m*n - m; i < m*n; i++){  // bottom edges (m*n - m = first char in the last string, it can be also looked as the last row in matrix)
        if(coast[i] == '1')
            edgeCoast++;
    }

    for(int i = 0; i <m*n; i+=m){   // left side edges (first column in matrix)
        if(coast[i] == '1')
            edgeCoast++;
    }

    for(int i = m-1; i < m*n; i+=m){ // right side edges (last column in matrix)
        if(coast[i] == '1')
            edgeCoast++;
    }

    return edgeCoast;
}

int topToBottomCount(int n, int m, char *coast){
    int coasts = 0;
    for(int i = 0; i < m*n - m; i++){   // we start from first char in "matrix", and move to the (m*n - m = 2nd last "row")
        if(coast[i] ^ coast[i+m])   // we are checking if zero is placed above one or via versa
            coasts++;
    }

    return coasts;
}

int leftToRightCount(int n, int m, char* coast){
    int coasts = 0;
    int p = m-1;
    for(int i = 0; i < n*m; i++){       // we start from the first charr, and we are going trough whole matrix, but the last column
        if(i == p){                     // p = m - 1 (last char in first row)
            p+=m;                       // p+=m (last char in next column, and so on)
            continue;                   // we move to next iteration
        }

        if(i == m*n - 1)                //if we are at last char in matrix, we break out from loop
            break;

        if(coast[i] ^ coast[i+1])
            coasts++;
    }

    return coasts;
}

int removingInsides(int n, int m, char* coast){ // Lakes and islands in lakes are not contributing to the sea coast. we are checking if they exist.
    int innerCoasts = 0;
    for(int i = m + 1; i < n*m - m - 1; i ++){
        if( coast[i] == '0' && coast[i] ^ coast[i-1] && coast[i] ^ coast[i+1] && coast[i] ^ coast[i-m] && coast[i] ^ coast[i+m]) // char has to be 0, and to hist left, right, above and under there has to be 1
            innerCoasts++;
    }

    return innerCoasts * 4; // *4 because we added 4 coasts before for each island.
}

Answer 1

我尝试使用 GCC C++ 编译器 (4.9.2) 编译您的代码。它编译得很好，我使用 the link you provided 中的示例问题对其进行了测试.它吐出正确答案。

但是，当我尝试使用 GCC C 编译器(也是 v 4.9.2)进行编译时，它失败并显示 'for' loop initial declarations are only allowed in C99 or C11 mode，由 this SO question 解释.我认为你的作业是使用 C 编译器评分的，由于这个错误，你的程序编译失败了。