Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E

Take the Land

Input : standard input

Output : standard output

 

The poor man went to the King and said, “Lord, I cannot maintain my family. Please give me some wealth so that I can survive with my wife and children.” The King replied, “I shall grant you a piece of land so that you can cultivate and grow food for your family. In the southern part of the Kingdom there is a rectangular forest. Trees have been planted there at regular intervals. Some of the trees have been cut for use. You are allowed to take any rectangular piece of land that does not contain any tree. You need not go to the forest to select the piece of land. I have a map containing 1s at places where there is a tree and 0s at points where the tree has been cut.”

 

Help the poor man to find out the largest piece of land. Area of the land is measured in units of number of trees that were there. Your program should take a matrix of 1s and 0s as input and output the area of the largest rectangular piece of land that contain no tree. Be careful about the efficiency of your program.

 

Input

The input file may contain multiple test cases. The first line of each test case contains two integers M and N  (1<=M,N<=100) giving the number of rows and columns in the matrix that follows. Each of the next M lines contains N symbols (either 0 or 1). Two consecutive symbols in a line will be separated by a single space. The input terminates with two zeros for M and N.

 

Output

For each test case in the input print a line giving the area (in terms of the number of trees were there) of the largest rectangular piece of land containing no tree.

 

Sample Input

6 7
0 1 1 0 1 1 0
0 0 0 0 0 1 0
1 0 0 0 0 0 1
0 1 0 0 0 0 1
1 1 0 0 0 1 0
1 1 0 1 1 0 0
0 0

Sample Output

12

---

做法 : DP
時間複雜度 O(N^2*M)

#include <stdio.h>

int main() {
    int m, n;
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0 && m == 0)
            break;
        int i, j, k, map[100][100];
        for(i = 0; i < n; i++)
            for(j = 0; j < m; j++)
                scanf("%d", &map[i][j]);
        int max = 0, tmp, length, width;
        for(i = 0; i < n; i++) {
            int sum[100] = {};
            for(j = i; j < n; j++) {
                for(k = 0; k < m; k++) {
                    sum[k] += !map[j][k];
                    if(k == 0 || tmp != length*width)
                        tmp = 0, length = 0;
                    length++, width = j-i+1;
                    tmp += sum[k];
                    if(tmp == length*width)
                        max = max > tmp ? max : tmp;
                }
            }
        }
        printf("%d\n", max);
    }
    return 0;
}