A计划 hdu 2102

此题刚开始TLE,愿意是没有考虑通过传输层到下一层时还是#,或*....

然后又开始了悲剧的wa...

原因是我从某点走到'#'时。。时间没有加1.。。

BFS。。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <queue>
#include <algorithm>
#include <iostream>

using namespace std;

char mp1[12][12];
char mp2[12][12];
int T, M, N, C, a, b, c, d, f1, f2, f3;
int xx[]={0, 1, -1, 0};
int yy[]={1, 0, 0, -1};

struct node
{
    int x, y, flag, t;
}Node;

deque<node>q;


int judge( int x, int y)
{
    if (x < 1 || x > N || y < 1 || y > M)
        return 0;
    return 1;

}

void BFS(  )
{
    int i, j, x1, x2,y1, y2, flag, t;
    q.clear( );
    Node.x = a, Node.y = b, Node.flag = f3, Node.t = 0;
    q.push_back(Node);
    while( !q.empty( ) && !f2)
    {
        Node = q.front( );
        q.pop_front( );
        x1 = Node.x, y1 =Node.y, flag = Node.flag, t = Node.t;
        //printf("(%d,%d,%d)--%d\n",x1,y1,flag,t);
        if (x1 == c && y1 == d && Node.flag == f1 && t <= T) {
            f2 = 1;
            //printf("%d\n",t);
            break;
        }
        //如果在第1层
        if (!flag)
        {  // 四个方向
            for (i = 0; i < 4; i++) {
                x2 = x1 + xx[i];
                y2 = y1 + yy[i];
                if (judge(x2,y2) && mp1[x2][y2] != '*')  // 满足条件
                {        
                    if (mp1[x2][y2] != '#')  // 第一层这点没有时空传输机
                   {
                      mp1[x2][y2] = '*';  //走过的点标记下，防止下次再走
                      Node.t = 0;
                      Node.x = x2, Node.y = y2, Node.flag = 0, Node.t = t + 1; // 此点入队
                      if (x2 == c && y2 == d && 0 == f1 && Node.t <= T) {
                      f2 = 1;
                      //printf("%d\n",Node.t);
                      break;
                    }
                    q.push_back(Node);
                    }
                   else   //有时空传输机    
                    {   
                     if ( mp2[x2][y2] == '#' || mp2[x2][y2] == '*') {
                            mp2[x2][y2] = '*';
                            continue;   // 穿输机不能又穿到传输机和墙
                       }
                     Node.t = 0;
                     mp2[x2][y2] = '*';
                     Node.x = x2, Node.y = y2,Node.flag = 1, Node.t = t  + 1;
                     if (x2 == c && y2 == d && 1 == f1 && Node.t <= T) {
                        f2 = 1;
                        //printf("%d\n",Node.t);
                        break;
                     }
                     q.push_back(Node);
                }
            }
            }
        }
        else  //如果此点是在第二层
        {
            for (i = 0; i < 4; i++) {  //四个方向
                x2 = x1 + xx[i];
                y2 = y1 + yy[i];
            if (judge(x2,y2) && mp2[x2][y2] != '*')  // 满足条件
            {  if (mp2[x2][y2] != '#')
                {
                   mp2[x2][y2] = '*';
                   Node.t = 0;
                   Node.x = x2, Node.y = y2, Node.flag = 1, Node.t = t + 1;
                 if (x2 == c && y2 == d && 1 == f1 && Node.t <= T) {
                    f2 = 1;
                   // printf("%d\n",Node.t);
                    break;
                  }
                 q.push_back(Node);
             }
                else
                {
                     
                     if (mp1[x2][y2] == '#' || mp1[x2][y2] == '*') {
                            mp1[x2][y2] = '*';
                            continue;
                      }
                     mp1[x2][y2] = '*';
                     Node.t = 0;
                     Node.x = x2, Node.y = y2,Node.flag = 0, Node.t = t + 1;
                     if (x2 == c && y2 == d && 0 == f1 && Node.t <= T) {
                        f2 = 1;
                       //printf("%d\n",Node.t);
                        break;
                      }
                     q.push_back(Node);
                }
            }
            }

        }
    }
}


               

int main( )
{   
    int i, j, k;
   
    scanf("%d", &C);
    while ( C --)
    {   
        f1 = f2 = f3 = 0;
        scanf("%d%d%d",&N, &M, &T);
        for (i = 1; i <= N; i++)
            for(j = 1; j <= M; j++) { 
                cin>>mp1[i][j];
                if ( mp1[i][j] == 'S')
                {
                    a = i;
                    b = j;
                
                }
                if ( mp1[i][j] == 'P')
                {
                    c = i;
                    d = j;
                }
          }
        for (i = 1; i <= N; i++)
            for(j = 1; j <= M; j++) {
                cin>>mp2[i][j];
                if ( mp2[i][j] == 'S')
                {
                    a = i;
                    b = j;
                    f3 = 1;
                }
                if ( mp2[i][j] == 'P')
                {
                    c = i;
                    d = j;
                    f1 = 1;
                }
            }
            BFS( );
            if (f2 == 1)
                puts("YES");
            else
                puts("NO");
    }
    return 0;
}