NYOJ-58 最小步数 AC 分类： NYOJ 2014-01-22 22:01 217人阅读 评论(0) 收藏

#include<stdio.h>

void dfs(int step,int x,int y);

int d[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};

int mg[9][9]={1,1,1,1,1,1,1,1,1,
              1,0,0,1,0,0,1,0,1,
              1,0,0,1,1,0,0,0,1,
              1,0,1,0,1,1,0,1,1,
              1,0,0,0,0,1,0,0,1,
              1,1,0,1,0,1,0,0,1,
              1,1,0,1,0,1,0,0,1,
              1,1,0,1,0,0,0,0,1,
              1,1,1,1,1,1,1,1,1,};
               
int sx,sy,ex,ey,min=100000;

int main(){
	
	int n;
	
	scanf("%d",&n);
	
	while(n--){
	
		scanf("%d %d %d %d",&sx,&sy,&ex,&ey);
		
		if((sx == ex)&&(sy == ey)){
			printf("0\n");
			continue;
		}
		
		mg[sx][sy]=1;
		mg[ex][ey]=3;
	
		dfs(0,sx,sy);
			
		printf("%d\n",min);
	
		min=100000;
		mg[ex][ey]=0;
		mg[sx][sy]=0;
		
	}
	return 0;
}

void dfs(int step,int x,int y){
	int i;

	for(i = 0 ; i < 4; i++)
    {
        int k =x+d[i][0] , v = y+d[i][1] ;

        if(!mg[k][v]){
        	
        	mg[k-d[i][0]][v-d[i][1]] = 1;
        	
        	dfs(step+1 , k , v) ;
        	
        	mg[k-d[i][0]][v-d[i][1]] = 0;
        }
        
        if(mg[k][v] == 3)
            {
                if(step+1 < min)
                    min = step + 1;
            }
            
    }
}

终于，终于，终于，把这题目搞定了，居然只是0步的情况没有考虑到，哎

版权声明：本文为博主原创文章，未经博主允许不得转载。