Question

我试图解决一个问题，我在网站https://open.kattis.com/problems/coast找到了这个问题。 Tl; dr版本的问题是，对于给定的景观地图，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，我已经过测试，据我检查，它有效。我认为它没有编译，但如果是这样，那为什么呢？它完全适合我。

#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 ++编译器编译代码（4.9.2）。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。它吐出了正确的答案。

但是，当我尝试使用GCC C 编译器（也是v 4.9.2）进行编译时，它失败了'for' loop initial declarations are only allowed in C99 or C11 mode，由this SO question解释。我认为您的作业使用C编译器进行了评分，并且由于此错误导致程序编译失败。

海岸长度，kattis

1 个答案: