洛谷 P1126 机器人搬重物

题目传送门

解题思路:

一道bfs,但本题有几个坑点.

1.题目给的图并不是最终可以用的图,要先手动转换成格点图.

2.机器人有一个直径,则说明原题给的图的障碍的四个节点都不能走.

3.接2,图的四周不能走,因为走的话机器人的一部分会出图.

4.机器人只能左转或右转,向后转需要两个时间.

其实解决了坑点,代码还是很好写的.

AC代码:

#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>

using namespace std;

int n,m,a[54][54],x2,y2,x1,y11,tot;
int tox[4][3] = {-1,-2,-3,0,0,0,1,2,3,0,0,0}; 
int toy[4][3] = {0,0,0,-1,-2,-3,0,0,0,1,2,3};
bool vis[54][54][4],ok[4][4] = {{0,1,0,1},{1,0,1,0},{0,1,0,1},{1,0,1,0}};
//ok用来判断是不是向左或向右转 
char o;
map<char,int> pp;
map<int,char> dp;
struct kkk{
    int xx,yy,time;
    char way;
}e[126001];
queue<kkk> q;

inline bool bfs(int x,int y) {
    vis[x][y][pp[o]] = 1;
    e[++tot].xx = x;
    e[tot].yy = y;
    e[tot].way = o;
    e[tot].time = 0;
    q.push(e[tot]);
    while(!q.empty()) {
        kkk k = q.front();
        q.pop();
        if(k.xx == x2 && k.yy == y2) {
            printf("%d",k.time);
            return false;
        }
        for(int i = 0;i <= 3; i++) {
            if(i == pp[k.way]) {
                for(int j = 0;j <= 2; j++) {
                    int tx = k.xx + tox[i][j];
                    int ty = k.yy + toy[i][j];
                    if(tx < 1 || tx >= n || ty < 1 || ty >= m) break;
                    if(a[tx][ty] == 1) break;
                    if(!vis[tx][ty][i]) {
                        vis[tx][ty][i] = 1;
                        e[++tot].xx = tx;
                        e[tot].yy = ty;
                        e[tot].time = k.time + 1;
                        e[tot].way = k.way;
                        q.push(e[tot]);
                    }
                }
            }
            else {
                if(ok[pp[k.way]][i] == 1 && !vis[k.xx][k.yy][i]) {
                    vis[k.xx][k.yy][i] = 1;
                    e[++tot].way = dp[i];
                    e[tot].time = k.time + 1;
                    e[tot].xx = k.xx;
                    e[tot].yy = k.yy;
                    q.push(e[tot]);
                }
            }
        }
    }
    return true;
}

inline void chushimap() {
    pp['N'] = 0;
    pp['W'] = 1;
    pp['S'] = 2;
    pp['E'] = 3;
    dp[0] = 'N';
    dp[1] = 'W';
    dp[2] = 'S';
    dp[3] = 'E';
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; i++)
        for(int j = 1;j <= m; j++) {
            int ooo;
            scanf("%d",&ooo);
            if(ooo == 1) {
                a[i][j] = 1;
                a[i][j-1] = 1;
                a[i-1][j] = 1;
                a[i-1][j-1] = 1;
            }
        }
    scanf("%d%d%d%d",&x1,&y11,&x2,&y2);
    cin >> o;
    chushimap();
//    for(int i = 1;i <= n+1; i++) {//格点图生成器 
//        for(int j = 1;j <= m+ 1; j++)
//            cout << a[i][j] << ' ';
//        cout << endl;
//    }
    if(bfs(x1,y11))
        printf("-1");
    
    return 0;
}