深度搜索hdu1010

HDU1010(DFS+奇偶剪枝) #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,m,t,a,b,c,d; char map[10][10]; bool vis[10][10]; int dir[10][10] = {{0,1},{0,-1},{-1,0},{1,0}}; bool flag; void dfs(int x,int y,i. 阅读详情

题目的大概意思是一只狗能不能在规定时间到达出口。如果可以,就输出”YES”,否则输出”NO”

思路:一道深度搜索的题,如果直接暴力搜索会超时,所以需要剪枝。当起点和终点的最短距离(无视障碍物)比规定的时间大的时候肯定是满足不了的,这是其中一个剪枝。还有一个很关键的剪枝,我们判断图的每个点是奇数还是偶数,如果起点和终点都是奇数或者偶数,那么需要的时间(步数)应该是偶数步,否则是奇数步。
比如:
输入的图是4x4矩阵,假设起点是(1,1)
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
此时起点(x+y)是偶数,假设终点是(1,2),两者不同为奇偶,那么需要的时间只能是奇数,如果终点是(2,2),同为奇偶,则需要的时间只能是偶数。即起点的坐标相加判断奇偶,终点的坐标相加判断奇偶,两者判断是否同奇偶,然后再判断对应的时间的奇偶性就行了。
代码:

#include <iostream>
using namespace std;
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h>
char map[8][8];
int flag[8][8];
int flag1;
int n, m, t;
void dfs(int step, int x, int y)
{
    if(flag1 == 1)
    {
        return ;
    }
    if(step == t) //时间用完了
    {
        if(map[x][y] == 'D')  //达到终点
        {
            printf("YES\n");  
            flag1 = 1;      //flag1用来防止有多条路径输出多个"YES"的情况
        }
        return;
    }
    if(x >0 && x <= n && y > 0 && y <= m && flag[x][y] == 0 && (map[x][y] == '.'||map[x][y] == 'S'))  //可以走
    {
        flag[x][y] = 1;
        dfs(step+1, x-1, y);
        dfs(step+1, x+1, y);
        dfs(step+1, x, y-1);
        dfs(step+1, x, y+1);
        flag[x][y] = 0; 
    }
    else
        return;
}
int main()
{
    while(scanf("%d %d %d", &n, &m, &t)!=EOF)
    {
        if(n == 0 && m == 0 && t == 0)
            return 0;
        getchar();
        int x, y, x1, y1;
        flag1 = 0;
        memset(map, 0, sizeof(map));
        memset(flag, 0, sizeof(flag));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%c", &map[i][j]);
                if(map[i][j] == 'S')
                {
                    x = i;
                    y = j;
                }
                if(map[i][j] == 'D')
                {
                    x1 = i;
                    y1 = j;
                }
            }
            getchar();
        }
        if(abs(x-x1)+abs(y-y1)>t || (x+y+x1+y1+t)%2 == 1)            //两个剪枝
        {
            printf("NO\n");
            continue;
        }
        dfs(0, x, y);
        if(flag1 == 0)
        {
            printf("NO\n");
        }   
    }
    return 0;
}
C++ 算法篇 深度优先搜索(DFS) 图的遍历 图的遍历为从图中某一顶点出发访遍图中其余顶点,且使每一个顶点仅被访问一次的过程。 对于图的遍历,不想树那么简单,需要在遍历的过程中把访问过的顶点打上标记,以避免访问多次。具体办法是设置一个访问数组visited[n],n是图中顶点的个数,初始值为0,访问后设置为1。 对于图的遍历来说,通常有两种遍历方案:深度优先遍历和广度优先遍历。 深度优先遍历 深度优先遍历(Depth_Fi... 阅读详情

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