Light OJ 1055 BFS

題目大意：給一個圖，圖上有A,B,C三個小人，每移動一次，都是對三個小人進行同時移動，如果碰到'#'，或者其他人的時候就自動不走。

思路：BFS求最短路，但是要同時記錄三個人的狀態，幸虧這道題的圖只有10*10,3個人的狀態也就10^6，不至於超內存，否則就要狀態壓縮了。

在我的思路里，BFS的時候判斷不走的條件是：沒有碰到'#‘，或者超出邊界。所以會出現碰到其他人繼續走的情況，wa了兩次。後來我的想法就是：記錄一下每個人在走每一步的時候是不是有移動了，然後如果移動過了，是不是會碰到其他人。這一步操作的判斷需要做兩次，因爲可能出現這種情況：C已經走到牆邊不能走，B在C的左邊，A在B的左邊。A先判斷自己和B的值是否重合，因爲B還沒判斷，所以A還是會重合到B，然後B判斷會重合C，保持狀態不動。在這種情況下，B的位置沒有問題，但是A會和B重合，所以判斷需要做兩次。

代碼效率49MS。

 

Code:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;

int n;
char a[20][20];
struct node
{
       int x, y;
}A, B, C;

struct status
{
       int ax, ay, bx, by, cx, cy, ti;
};

int times[11][11][11][11][11][11];
int xx[4] = {1, 0, -1, 0};
int yy[4] = {0, 1, 0, -1};

bool isok(int x, int y)
{
     return (x >= 0 && y >= 0 && x < n && y < n);
}

int bfs()
{
    queue<status> q;
    status top, tmp;
    top.ax = A.x, top.ay = A.y, top.bx = B.x, top.by = B.y, top.cx = C.x, top.cy = C.y;
    top.ti = 0;
    q.push(top);
    while(!q.empty())
    {
                int i;
                top = q.front();
                q.pop();
                for(i = 0; i < 4; i ++)
                {
                      int pax = xx[i] + top.ax;
                      int pay = yy[i] + top.ay;
                      int pbx = xx[i] + top.bx;
                      int pby = yy[i] + top.by;
                      int pcx = xx[i] + top.cx;
                      int pcy = yy[i] + top.cy;
                      bool oka = false, okb = false, okc = false;
                      if(!isok(pax, pay) || a[pax][pay] == '#')
                      {
                                    pax = top.ax, pay = top.ay;
                                    oka = true;
                      }
                      if(!isok(pbx, pby) || a[pbx][pby] == '#')
                      {
                                    pbx = top.bx, pby = top.by;
                                    okb = true;
                      }
                      if(!isok(pcx, pcy) || a[pcx][pcy] == '#')
                      {
                                    pcx = top.cx, pcy = top.cy;
                                    okc = true;
                      }
                      if((!oka && okc && pax == pcx && pay == pcy) || (!oka && okb && pax == pbx && pay == pby))
                      {
                              pax = top.ax, pay = top.ay;
                              oka = true;
                      }
                      if((!okb && okc && pbx == pcx && pby == pcy) || (oka && !okb && pax == pbx && pay == pby))
                      {
                              pbx = top.bx, pby = top.by;
                              okb = true;
                      }
                      if((oka && !okc && pax == pcx && pay == pcy) || (!okc && okb && pcx == pbx && pcy == pby))
                      {
                              pcx = top.cx, pcy = top.cy;
                              okc = true;
                      }
                      if((!oka && okc && pax == pcx && pay == pcy) || (!oka && okb && pax == pbx && pay == pby))
                      {
                              pax = top.ax, pay = top.ay;
                              oka = true;
                      }
                      if((!okb && okc && pbx == pcx && pby == pcy) || (oka && !okb && pax == pbx && pay == pby))
                      {
                              pbx = top.bx, pby = top.by;
                              okb = true;
                      }
                      if((oka && !okc && pax == pcx && pay == pcy) || (!okc && okb && pcx == pbx && pcy == pby))
                      {
                              pcx = top.cx, pcy = top.cy;
                              okc = true;
                      }
                      if(times[pax][pay][pbx][pby][pcx][pcy] == 0 || times[pax][pay][pbx][pby][pcx][pcy] > top.ti + 1)
                      {
                                                             times[pax][pay][pbx][pby][pcx][pcy] = tmp.ti = top.ti + 1;
                                                             tmp.ax = pax, tmp.ay = pay, tmp.bx = pbx, tmp.by = pby, tmp.cx = pcx, tmp.cy = pcy;
                                                             q.push(tmp);
                                                             if(a[pax][pay] == 'X' && a[pbx][pby] == 'X' && a[pcx][pcy] == 'X') //&& (!(pax == pbx && pay == pby) && !(pax == pcx && pay == pcy) && !(pcx == pbx && pcy == pby)))
                                                             {
                                                                           // cout << pax << "  " << pay << "  " << pbx << " " << pby << " " << pcx << " " << pcy << endl;
                                                                            return tmp.ti;
                                                             }
                      }
                }
    }
    return -1;
}

int main()
{
    int cse, t = 1, i, j;
    scanf("%d", &cse);
    while(cse --)
    {
              scanf("%d", &n);
              for(i = 0; i < n; i ++)
                    scanf("%s", a[i]);
              for(i = 0; i < n; i ++)
                    for(j = 0; j < n; j ++)
                          if(a[i][j] == 'A')
                                     A.x = i, A.y = j;
                          else if(a[i][j] == 'B')
                               B.x = i, B.y = j;
                          else if(a[i][j] == 'C')
                               C.x = i, C.y = j;
              memset(times, 0, sizeof(times));
              int ans = bfs();
              printf("Case %d: ", t ++);
              if(ans != -1)
                     printf("%d\n", ans);
              else
                     printf("trapped\n");
    }
}