搜索例题

http://acm.sdibt.edu.cn/vjudge/contest/view.action?cid=1977#problem/A

广搜:

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

在一个8*8的棋盘中,a1代表第零行的第一个元素,每个棋子只能像象棋中“马”跳的方式跳,问从前面跳到后面需要最少几步(广搜);

代码如下:

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<algorithm>
using namespace std;
char a,b;
int a1,b1;
int mapp[10][10];
int bj[9][3]= {{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};
struct p
{
    char a;
    int a1;
    int step;
} x,y,neww;
queue<p>q;
int bfs()
{
    int i;
    x.a=a;
    x.a1=a1;
    x.step=0;
    while(q.size()!=0)
        q.pop();
    memset(mapp,0,sizeof(mapp));
    mapp[x.a-'a'][x.a1]=1;
    q.push(x);
    while(q.size()!=0)
    {
        y=q.front();
        q.pop();
        if(y.a==b&&y.a1==b1)
        {
            return y.step;
        }
        x=y;
        for(i=0; i<=7; i++)
        {
            int xx=x.a-'a'+bj[i][0];   
            int yy=x.a1+bj[i][1];
            if(xx>=0&&xx<=7&&yy>=1&&yy<=8&&mapp[xx][yy]==0)
            {
                neww.step=x.step+1;
                mapp[xx][yy]=1;
                neww.a=xx+'a';
                neww.a1=yy;
                q.push(neww);
            }
        }

    }
}
int main()
{
    while(~scanf("%c",&a))
    {
        scanf("%d",&a1);
        getchar();
        scanf("%c",&b);
        scanf("%d",&b1);
        getchar();
        printf("To get from %c%d to %c%d takes %d knight moves.\n",a,a1,b,b1,bfs());
    }
    return 0;
}

深搜:

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

45
59
6
13

#include <iostream>
#include<stdio.h>
#include<string.h>
int a,b;
char c[200][200];
int d[200][200];
using namespace std;
void l(int i,int j)
{
    if(i>=0&&j>=0&&i<b&&j<a)
    {
        if(d[i][j]==0&&c[i][j]=='.')
        {
            d[i][j]=1;
            l(i,j-1);
            l(i,j+1);
            l(i+1,j);
            l(i-1,j);
        }
    }
}
int main()
{
    int m,n,sum=0,i,j;
    while(~scanf("%d%d",&a,&b))
    {
        getchar();
        memset(d,0,sizeof(d));
        if(a==b&&a==0)
            break;
        sum=0;
        for(i=0; i<=b-1; i++)
        {
            for(j=0; j<=a-1; j++)
            {
                scanf("%c",&c[i][j]);
                if(c[i][j]=='@')
                {
                    m=i;
                    n=j;
                }
            }
            getchar();
        }
        c[m][n]='.';
        l(m,n);
        for(i=0; i<=b-1; i++)
            for(j=0; j<=a-1; j++)
                    sum+=d[i][j];
        printf("%d\n",sum);
    }


}

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转载自www.cnblogs.com/bhd123/p/9498138.html