HDU 1312 统计有多少个点

最新推荐文章于 2021-10-07 12:00:22 发布

原创最新推荐文章于 2021-10-07 12:00:22 发布 · 置顶 · 570 阅读

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#搜索

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本文探讨了使用广度优先搜索（BFS）和深度优先搜索（DFS）解决地图探索问题的方法，其中地图由黑色和红色方块组成，玩家只能在黑色方块上移动，目标是计算玩家能够到达的黑色方块总数。通过实例分析，详细解释了两种搜索算法的实现过程及应用场景。

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Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10858 Accepted Submission(s): 6797

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

题意：一幅地图，#不能走，@代表进去的点，问你有多少个点（.），@也算一个。

比较水的一道搜索题。

我写了两种代码。

BFS和DFS

#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
struct node
{
	int x,y;
}
f[333];
#define inf 0xfffffff
void bfs(int x,int y)
{
	int i;
	queue<node>q;
	node st,ed;
	st.x=x;
	st.y=y;
	q.push(st);
	while(!q.empty())
	{
		st=q.front();
		q.pop();
		for(i=0;i<4;i++)
		{
			ed.x=st.x+dir[i][0];
			ed.y=st.y+dir[i][1];
			if(ed.x>=n ||ed.y>=m ||ed.x<0 ||ed.y<0 ||map[ed.x][ed.y]=='#' ||map[ed.x][ed.y]=='@')
				continue;
			map[ed.x][ed.y]='@';
			sum++;
			q.push(ed);
		}
	}
}
int main()
{
	int i,j,x,y;
	while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
	{
		for(i=0;i<n;i++)
			scanf("%s",map[i]);
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='@')
				{
					x=i;
					y=j;		 
				}
			}
			sum=1;
			bfs(x,y);
			printf("%d\n",sum);
	}
	return 0;
}
//广搜








#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
#define inf 0xfffffff
void dfs(int x,int y)
{
    int sx,sy,i;
    sum++;
    for(i=0;i<4;i++)
    {
        sx=x+dir[i][0];
        sy=y+dir[i][1];
        if(sx>=n ||sy>=m ||sx<0 ||sy<0 ||map[sx][sy]=='#' ||map[sx][sy]=='@')
            continue;
          map[sx][sy]='@';
              dfs(sx,sy);
    } 
}
int main()
{
    int i,j,x,y;
    while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
    {
        for(i=0;i<n;i++)
            scanf("%s",map[i]);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='@')
                {
                    x=i;
                    y=j;         
                }
            }
            sum=0;
            dfs(x,y);
            printf("%d\n",sum);
    }
    return 0;
}
//深搜
//深搜和广搜都是只能讲满足的点放入继续搜索，所以搜过的点要把它变成不满足的点，满足的点可以继续再搜。这就是本题的关键