寻找一个图中独立子图的个数

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最新推荐文章于 2024-02-21 16:14:17 发布 · 1.6k 阅读

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#算法 #地图

#include <stdio.h>

int map[51][51] = 
{
	{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
	{0, 1, 2, 1, 0, 0, 0, 0, 0, 2, 3},
	{0, 3, 0, 2, 0, 1, 2, 1, 0, 1, 2},
	{0, 4, 0, 1, 0, 1, 2, 3, 2, 0, 1},
	{0, 3, 2, 0, 0, 0, 1, 2, 4, 0, 0},
	{0, 0, 0, 0, 0, 0, 0, 1, 5, 3, 0},
	{0, 0, 1, 2, 1, 0, 1, 5, 4, 3, 0},
	{0, 0, 1, 2, 3, 1, 3, 6, 2, 1, 0},
	{0, 0, 0, 3, 4, 8, 9, 7, 5, 0, 0},
	{0, 0, 0, 0, 3, 7, 8, 6, 0, 1, 2},
	{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0}
};
int book[51][51] = {0};
int sum = 0;
int next[4][2] = {
  
  {0, 1}, {1, 0}, {0, -1}, {-1, 0}};
void dfs(int x, int y,int color)
{
	int k;
	int tx, ty;
	map[x][y] = color;
	for (k = 0; k <= 3; k++)
		{
			tx = x + next[k][0];
			ty = y + next[k][1];

			if (tx < 1 || tx > 10 || ty < 1 || ty > 10)
				continue;

			if (map[tx][ty] > 0 && book[tx][ty] == 0)
			{
				sum++;
				book[tx][ty] = 1;
				dfs(tx, ty, color);
				//book[tx][ty] = 0;
			}
		}
	return;
}
int main()
{
	int i,j;
	int