宝岛探险 BFS DFS

最新推荐文章于 2022-09-07 16:56:29 发布

最新推荐文章于 2022-09-07 16:56:29 发布 · 1.2k 阅读

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本文通过广度优先搜索(BFS)和深度优先搜索(DFS)两种算法计算从指定位置出发的岛屿面积，并采用着色法统计地图上的岛屿总数。代码示例清晰展示了不同算法的应用。

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1.计算岛屿面积

1.1 BFS

#include <stdio.h>
int a[10][10] = {
    {1,2,1,0,0,0,0,0,2,3},
    {3,0,2,0,1,2,1,0,1,2},
    {4,0,1,0,1,2,3,2,0,1},
    {3,2,0,0,0,1,2,4,0,1},
    {0,0,0,0,0,0,1,5,3,0},
    {0,1,2,1,0,1,5,4,3,0},
    {0,1,2,3,1,3,6,2,1,0},
    {0,0,3,4,8,9,7,5,0,0},
    {0,0,0,3,7,8,6,0,1,2},
    {0,0,0,0,0,0,0,0,1,0}
    };
/*
0表示海洋，1-9表示陆地的海拔，降落在(6,8)处，计算出该岛的面积（有多少个格子）。
即从(6,8)处进行广度优先搜索, sum ++
*/
struct note{
    int x;
    int y;
};
int main(){
    struct note que[101];
    int head,tail;
    int book[10][10]={0};
    int startx = 6, starty = 8;
    int sum=0;
    int next[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};//右下左上

    //初始化队列
    head = 1;
    tail = 1;
    que[tail].x = startx;
    que[tail].y = starty;
    book[startx][starty] = 1;
    tail++;
    sum++;
    while(head < tail){
        for(k = 0; k <= 3; k++){
            tx = que[head].x + next[k][0];
            ty = que[head].y + next[k][1];
            if(tx < 0 || tx > 9 || ty < 0 || ty > 9){
            continue;
            }
            if(a[tx][ty] > 0 && book[tx][ty] == 0){
                book[tx][ty] = 1;
                sum++;
                que[tail].x = tx;
                que[tail].y = ty;
                tail++;
            }

        }
        head++;
    }
    printf("该岛屿面积为%d.\n",sum);
    getchar();
    return 0;

}

1.2 DFS

#include <stdio.h>
int a[10][10] = {
    {1,2,1,0,0,0,0,0,2,3},
    {3,0,2,0,1,2,1,0,1,2},
    {4,0,1,0,1,2,3,2,0,1},
    {3,2,0,0,0,1,2,4,0,1},
    {0,0,0,0,0,0,1,5,3,0},
    {0,1,2,1,0,1,5,4,3,0},
    {0,1,2,3,1,3,6,2,1,0},
    {0,0,3,4,8,9,7,5,0,0},
    {0,0,0,3,7,8,6,0,1,2},
    {0,0,0,0,0,0,0,0,1,0}
    };
int next[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};//右下左上
/*
0表示海洋，1-9表示陆地的海拔，降落在(6,8)处，计算出该岛的面积（有多少个格子）。
即从(6,8)处进行深度优先搜索, sum ++
*/
int book[10][10] = {0};
int sum;
void dfs(int x,int y){
    int k,tx,ty;

    for(k=0;k<=3;k++){
        tx = x + next[k][0];
        ty = y + next[k][1];
        if(tx < 0 || tx > 9 || ty < 0 || ty > 9){
            continue;
        }
        if(a[tx][ty] > 0 && book[tx][ty] == 0){
            sum ++;
            book[tx][ty] = 1;
            dfs(tx,ty);
        }
    }
    return ;
}

int main(){
    int i,j,startx = 6,start = 8;
    book[startx][starty] = 1;
    sum = 1;
    dfs(startx.starty);
    printf("%d\n",sum);
    getchar();
    return 0;

}

2.一共有几个岛（独立子图个数）

着色法

#include <stdio.h>
int a[10][10] = {
    {1,2,1,0,0,0,0,0,2,3},
    {3,0,2,0,1,2,1,0,1,2},
    {4,0,1,0,1,2,3,2,0,1},
    {3,2,0,0,0,1,2,4,0,1},
    {0,0,0,0,0,0,1,5,3,0},
    {0,1,2,1,0,1,5,4,3,0},
    {0,1,2,3,1,3,6,2,1,0},
    {0,0,3,4,8,9,7,5,0,0},
    {0,0,0,3,7,8,6,0,1,2},
    {0,0,0,0,0,0,0,0,1,0}
    };

/*
0表示海洋，1-9表示陆地的海拔。求有几个岛屿。
*/
int next[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};//右下左上
int book[10][10] = {0};
int sum;

void dfs(int x,int y, int color){
    int k, tx, ty;
    a[x][y] = color;
    for(k = 0; k <= 3; k++){
        tx = x + next[k][0];
        ty = y + next[k][1];
        if(tx < 0 || tx > 9 || ty < 0 || ty > 9){
            continue;
        }
        if(a[tx][ty] > 0 && book[tx][ty] == 0){
            book[tx][ty] = 1;
            sum++;
            dfs(tx,ty,color);
        }
    }
    return ;
}

int main(){
    int i,j,num = 0;
    for(i=0; i<=9;i++){
        for(j=0;j<=9;j++){
            if(a[i][j] > 0){
                num--;
                book[i][j] = 1;
                dfs(i,j,num);
            }
        }
    }
    //输出着色后的地图
    for(i=0; i<=9;i++){
        for(j=0;j<=9;j++){
            printf("%3d",a[i][j]);
        }
        printf("\n");
    }
    printf("共有%d个小岛。",-num);
    getchar();
    return 0;
}