NYOJ-58 最小步数 AC

来源：互联网发布：tensorflow 1.3 whl 编辑：程序博客网时间：2024/05/27 20:32

#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

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