BFS algorithm Notes

• To 注意题目中给的数据的范围, to determine the size of an array, or else find the cause of the timeout to no avail

step

1. The initial point is added to the team next to the search
2. while (head <tail) to start the cycle
3. Each cycle is determined around 4 points for(int i=0;i<4;i++)
4. To determine whether cross-border and came to decide whether to join the ranks of go
5. After the cycle is completed for each, to be in the ranks of the head, in a increasehead++

example

• The largest islands

Description

An ocean area has a number of islands, find the largest island in the area.
With a 0 and contains some non-empty one-dimensional array to represent this area, which represents an area of land 1, 0 represents water. You can assume that the four edges of the two-dimensional matrix are surrounded by water.
Find a given two-dimensional array in the largest island in the area. (If there are no islands, the area is returned to 0.)

Input

The first line of two integers nm (n <= 100 m < = 100)
the next input two-dimensional array of n rows and m columns.

Output

The largest island of the output area (if not the island, the area returns to 0).

Sample Input

4 5
1 0 0 1 1
1 0 1 1 0
0 1 0 0 0
1 1 0 0 1

Sample Output

4

• answer

#include <iostream>
using namespace std;

struct note{
    int x;
    int y;
};
int nx[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};

int main()
{
    // 超时原因可能是数组太小
    int n,m,dx,dy,mm=0;
    int data[101][101];
    note que[10000];
    int head,tail;
    int area;
    cin >> n >> m;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            scanf("%d",&data[i][j]);
        }
    }
    // 搜索
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(data[i][j]==1){
                data[i][j] = 0;
                head=0,tail=0;
                que[tail].x = i;
                que[tail].y = j;
                tail++;
                area = 1;
                while(head < tail){
                    for(int k=0;k<4;k++){
                        dx = que[head].x + nx[k][0];
                        dy = que[head].y + nx[k][1];
                        if(data[dx][dy]==0) continue;
                        if(dx >=0 && dy>=0 && dx < n && dy <m){
                            data[dx][dy] = 0;
                            que[tail].x = dx;
                            que[tail].y = dy;
                            tail++;
                            area++;
                        }
                    }
                    head++;
                }
                if(area>mm){
                    mm = area;
                }
            }
        }
    }
    cout << mm << endl;
    return 0;
}