深搜hdu1312

来源：互联网发布：有什么金融软件编辑：程序博客网时间：2024/06/06 00:55

一般来说，广搜常用于找单一的最短路线，或者是规模小的路径搜索，它的特点是"搜到就是最优解"，而深搜用于找多个解或者是"步数已知（好比3步就必需达到前提）"的标题，它的空间效率高，然则找到的不必定是最优解，必需记实并完成全数搜索，故一般情况下，深搜需要很是高效的剪枝（优化）.（剪枝真真的要命啊）

像搜索最短路径这些的很显著若是用广搜，因为广搜的特征就是一层一层往下搜的，保证当前搜到的都是最优解，当然，最短路径只是一方面的操作，像什么起码状态转换也是可以操作的。
深搜就是优先搜索一棵子树，然后是另一棵，它和广搜对比，有着内存需要相对较少的所长，八皇后标题就是典范楷模的操作，这类标题很显著是不能用广搜往解决的。或者像图论里面的找圈的算法，数的前序中序后序遍历等，都是深搜。下面就拿出一道深搜的题目练练手。

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10104 Accepted Submission(s): 6301

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

思路很好找啊，直接找完图中所有点就好~

话不多说直接上代码

#include<stdio.h>int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};//四个方向int m,n,bx,by,cnt;char map[22][22];void dfs(int x,int y)//深搜求所有符合要求的点{    int i;cnt++;    for(i=0;i<4;i++)    {        int fx=x+dir[i][0];        int fy=y+dir[i][1];        if(fx>=0&&fx<n&&fy>=0&&fy<m&&map[fx][fy]=='.')        {            map[fx][fy]='#' ;//找到一个黑块标记一下下次不可以再走            dfs(fx,fy);        }    }}int main(){    int i,j;    //freopen("int.txt","r",stdin);    while(scanf("%d%d",&m,&n)!=EOF)    {        getchar();        if(m==0&&n==0)break;        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                scanf("%c",&map[i][j]);                if(map[i][j]=='@'){bx=i;by=j;}            }            getchar();        }        cnt=0;map[bx][by]='#';        dfs(bx,by);        printf("%d\n",cnt);    }    return 0;}

一遍ac是不是很爽啊~那么我要找一点有难度的搜索喽

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