HDU2102 A计划

解题思路：一道简单题，却WA了十几发，犯几个低级错误。还是不能急躁，

　　　     内心要平静，具体分析见代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 15;
char mapp[maxn][maxn][2];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int n, m, t, c;

struct node{
    int x, y;
    int f;
    int step;
}s, e;

queue<node> q;

int bfs()
{
    while(!q.empty()) q.pop(); //没加这步WA了好多发
    q.push(s);

    while(!q.empty())
    {
        s = q.front(); q.pop();
        //debug时可打印路径
        //printf("s.x = %d, s.y = %d, s.step = %d, s.flag = %d
", s.x, s.y, s.step, s.flag);


        for(int i = 0; i < 4; i++)
        {
            e.x = s.x + dir[i][0];
            e.y = s.y + dir[i][1];
            e.step = s.step + 1;
            e.f = s.f;

            if(e.step > t || mapp[e.x][e.y][e.f] == '*') continue;//超过时间或不能走
            if(mapp[e.x][e.y][e.f] == '#')
            {
                mapp[e.x][e.y][e.f] = '*'; //标记为已走过
                e.f = 1 - e.f; //穿越到另一层
                //另一层若为#或*则标记为不能走
                if(mapp[e.x][e.y][e.f] == '#' || mapp[e.x][e.y][e.f] == '*')
                {
                    mapp[e.x][e.y][e.f] = '*';
                    continue; //必不可少
                }
            }

            if(mapp[e.x][e.y][e.f] == 'P') return 1; //找到公主
            mapp[e.x][e.y][e.f] = '*'; //标记为已走过
            q.push(e);
        }
    }
    return 0; //规定时间没找到
}

int main()
{
    scanf("%d", &c);
    while(c --)
    {
        memset(mapp, '*', sizeof(mapp));
        scanf("%d %d %d", &n, &m, &t);

        for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++)
        scanf(" %c", &mapp[i][j][0]);

        for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++)
        scanf(" %c", &mapp[i][j][1]);

        s.x = 1, s.y = 1, s.step = 0, s.f = 0;
        if(bfs()) printf("YES
");
        else printf("NO
");
    }
    return 0;
}