题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2102

题解：

BFS+剪枝

1.题意不是很清楚，不太清楚S是否一定是（0,0,0），所以这里做了一次查找；

2.对于传送门的问题，如果两个传送门相对，那么骑士一定会被传送门华丽丽地玩死了；

3.剪枝，这里的剪枝由于两层之间传送不消耗时间，所以我们只需要求：T(当前位置)+DIS(公主位置-当前位置)<=T才能继续进行，否则就要被剪掉

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define Lowbit(x) ((x)&(-(x)))
#define ll long long
#define mp make_pair
#define ff first
#define ss second
#define pb push_back
#define mod 10000007
//#define LOCAL
#define MAXN 100010
#define INF 1e9
#define Max 100010

int ty[]={0,1,0,-1};
int tz[]={1,0,-1,0};

char str[2][15][15];
bool visit[2][15][15];
int N,M,T;

typedef struct node{
    int x,y,z;
    int cost;
}NODE;

bool check(int &x,int &y,int &z){
    if(y<0||y>=N||z<0||z>=M||str[x][y][z]=='*'||visit[x][y][z])
        return false;
    visit[x][y][z] = true;
    if(str[x][y][z]=='#')
        if(str[x^1][y][z]=='*'||str[x^1][y][z]=='#')
            return false;
        else
            x^=1;   //传送门进行层传递
    visit[x][y][z] = true;

    return true;
}

int main(){
    int C;
    scanf("%d",&C);
    while(C--){
        scanf("%d%d%d", &N, &M, &T);
        int i,j,k;
        for(i=0; i<2; ++i){
            for(j=0; j<N; ++j)
                scanf("%s", str[i][j]);
        }

        NODE now;
        NODE target;
        //查找'S'\'P'点的位置
        for(i=0; i<2; ++i){
            for(j=0; j<N; ++j){
                for(k=0; k<M; ++k){
                    if(str[i][j][k]=='S'){
                        now.x = i;
                        now.y = j;
                        now.z = k;
                    }
                    if(str[i][j][k]=='P'){
                        target.x = i;
                        target.y = j;
                        target.z = k;
                    }
                }
            }
        }

        //BFS
        memset(visit, false, sizeof(visit));
        now.cost = 0;
        queue<NODE> que;
        que.push(now);
        visit[now.x][now.y][now.z] = true;
        bool flag = false;
        while(que.size()){
            now = que.front();
            que.pop();

            if(now.cost>T)  break;
            if(now.x==target.x&&now.y==target.y&&now.z==target.z){
                flag = true;
                break;
            }
            NODE next;
            for(i=0; i<4; ++i){
                next.x = now.x;
                next.y = now.y+ty[i];
                next.z = now.z+tz[i];
                next.cost = now.cost+1;
                if(check(next.x,next.y,next.z)){
                    //剪枝
                    if(next.cost+
                       abs(target.y-next.y)+abs(target.z-next.z)>T)
                        continue;
                    que.push(next);
                }
            }
        }
        printf("%s\n", flag?"YES":"NO");
    }
    return 0;
}

HDU 2102 A计划