宝岛冒险
题目
小哼通过秘密方法得到了一张不完整的钓鱼岛航拍地图。钓鱼岛由一个主岛和一些附属岛屿组成,小哼决定去钓鱼岛冒险。下面这个 10 * 10 的二维码矩阵就是钓鱼岛的航拍地图。图中数字代表海拔,0 表示海洋,1~9 表示陆地。小哼的飞机将会在(6,8)处,现在需要计算出小哼落地所在岛屿的面积(即有多少个格子)。
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 0
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
计算小岛面积
#include <stdio.h>
#define NUM 10
int a[NUM][NUM] = {
{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, 0},
{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 book[NUM][NUM];
int direction[4][2] = {
{-1, 0},
{0, 1},
{1, 0},
{0, -1}
};
int main(void) {
int i, j, tx, ty, area = 0;
for (i = 0; i < NUM; i++)
for (j = 0; j < NUM; j++)
book[i][j] = 0;
// 初始化队列
struct node {
int x;
int y;
} que[1000];
int head = 0;
int tail = 0;
// 初始化位置
que[tail].x = 6;
que[tail].y = 8;
tail++;
book[6][8] = 1;
area++;
while (head < tail) {
for (i = 0; i < 4; i++) {
tx = que[head].x + direction[i][0];
ty = que[head].y + direction[i][1];
if (tx >= 0 && tx < NUM && ty >= 0 && ty < NUM) {
if (a[tx][ty] > 0 && book[tx][ty] == 0) {
book[tx][ty] = 1;
a[tx][ty] = -1;
que[tail].x = tx;
que[tail].y = ty;
tail++;
area++;
}
}
}
head++;
}
printf("%d\n", area);
for (i = 0; i < NUM; i++) {
for (j = 0; j < NUM; j++) {
printf("%2d ", a[i][j]);
}
printf("\n");
}
return 0;
}
用符号表示路过的小岛
#include <stdio.h>
#define NUM 10
int a[NUM][NUM] = {
{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, 0},
{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 book[NUM][NUM];
int sum = 0;
int direction[4][2] = {
{-1, 0},
{0, 1},
{1, 0},
{0, -1}
};
void dfs(int x, int y, int color) {
int i, tx, ty;
// 边界
a[x][y] = color;
// 尝试每一步
for (i = 0; i < 4; i++) {
tx = x + direction[i][0];
ty = y + direction[i][1];
if (tx >= 0 && tx < NUM && ty >= 0 && ty < NUM) {
if (a[tx][ty] > 0 && book[tx][ty] == 0) {
book[tx][ty] = 1;
sum++;
// 下一步
dfs(tx, ty, color);
}
}
}
}
int main(void) {
int i, j;
dfs(6, 8, -1);
printf("%d\n", sum);
for (i = 0; i < NUM; i++) {
for (j = 0; j < NUM; j++) {
printf("%2d ", a[i][j]);
}
printf("\n");
}
return 0;
}
记录地图有多少个小岛
#include <stdio.h>
#define NUM 10
int a[NUM][NUM] = {
{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, 0},
{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 book[NUM][NUM];
int direction[4][2] = {
{-1, 0},
{0, 1},
{1, 0},
{0, -1}
};
void dfs(int x, int y, int color) {
int i, tx, ty;
// 边界
a[x][y] = color;
// 尝试每一步
for (i = 0; i < 4; i++) {
tx = x + direction[i][0];
ty = y + direction[i][1];
if (tx >= 0 && tx < NUM && ty >= 0 && ty < NUM) {
printf("%d %d\n", tx, ty);
if (a[tx][ty] > 0 && book[tx][ty] == 0) {
book[tx][ty] = 1;
// 下一步
dfs(tx, ty, color);
}
}
}
}
int main(void) {
int i, j, num = 0;
for (i = 0; i < NUM; i++) {
for (j = 0; j < NUM; j++) {
if (a[i][j] > 0 && book[i][j] == 0) {
num--;
book[i][j] = 1;
dfs(i, j, num);
}
}
}
// dfs(0, 0, -1);
for (i = 0; i < NUM; i++) {
for (j = 0; j < NUM; j++) {
printf("%2d ", a[i][j]);
}
printf("\n");
}
printf("%d\n", -num);
return 0;
}
这一个算法就是鼎鼎大名的 FloodFill 漫水填充法(也称种子填充法)。常用于图像分割、物体识别等等。
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