搜索算法集锦 Ⅰ

搜索：带有方向性的枚举（自我理解）
分类：广度优先搜索(BFS)和深度优先搜索(DFS)（含有回溯）

[hdu1312：] (http://acm.hdu.edu.cn/showproblem.php?pid=1312)
题目描述：
Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13171 Accepted Submission(s): 8163

Problem Description
There is a rectangular（矩形） room, covered with square tiles. Each tile（地砖） 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（临近的） 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（最初的） tile (including itself).

Sample Input
6 9
….#.
…..#
……
……
……
……
……

@…

.#..#.
11 9
.#………
.#.#######.
.#.#…..#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#…….#.
.#########.
………..
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..

.

…@…

.

..#.#..
..#.#..
0 0

Sample Output
45
59
6
13

【言简意赅】：
计算从’@’开始能经过的所有’.’的数量，路径可重复，但总数不重复记录

广搜代码实现：

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

const int maxn=30;
char a[maxn][maxn];//存图数组
bool b[maxn][maxn];//判重数组
int qx[maxn*maxn],qy[maxn*maxn];
int vb[4][2]= {{-1,0},{1,0},{0,1},{0,-1}};//上下左右四个方向
int ans,m,n;
void bfs(int x,int y)
{
    int l=0,r=0;
    qx[r]=x;
    qy[r++]=y;
    //qx[r]=x;qy[r]=y;r++;
    b[x][y]=1;
    ans++;
    while(l<r)
    {
        int cx=qx[l];
        int cy=qy[l++];
        //int cx=qx[l];int cy=qy[l];l++;
        for(int i=0; i<4; i++)
        {
            int nx=cx+vb[i][0];
            int ny=cy+vb[i][1];
            if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!b[nx][ny]&&a[nx][ny]!='#')
            {
                ans++;
                b[nx][ny]=1;//状态标记
                qx[r]=nx;qy[r++]=ny;//注意！此处是状态转移
                //qx[r]=nx;qy[r]=ny;r++;
            }
        }
    }
}
int main()
{
    int x,y;
    while(scanf("%d%d",&m,&n)!=EOF)//此题是应用广搜
    {
        getchar();
        if(m==0&&n==0)
        break;
        memset(a,0,sizeof(a));
        memset(b,false,sizeof(b));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                cin>>a[i][j];
                if(a[i][j]=='@')//由于只有一个'@'，即可以在输入时就将其下标记录下来
                {
                    x=i;
                    y=j;
                }
            }
        }
        ans=0;
        bfs(x,y);
        cout<<ans<<endl;
    }
    return 0;
}

深搜代码实现：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=30;
int n,m;
int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
char map[maxn][maxn];
bool vis[maxn][maxn];
int ans;
void dfs(int x,int y)
{
    vis[x][y]++;
    ans++;
    for(int i=0; i<4; i++)//在可选的4个方向上依次实现搜索
    {
        int nx=x+dir[i][0];
        int ny=y+dir[i][1];
        if(nx<1||nx>n) continue;
        if(ny<1||ny>m) continue;
        if(vis[nx][ny]||map[nx][ny]=='#') continue;
        dfs(nx,ny);//递归实现深度优先搜索
    }
}
int main()
{
    int ca,a,b,cas=1;
    cin>>ca;
    while(ca--)
    {
        cin>>m>>n;
        int sx,sy;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@') sx=i,sy=j;
            }
        memset(vis,0,sizeof(vis));
        ans=0;
        dfs(sx,sy);
        cout<<"Case "<<cas++<<": "<<ans<<endl;
    }
    return 0;
}