HDU 2102 A计划

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

题解：

BFS+剪枝

1.题意不是非常清楚，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，假设两个传送门相对，那么骑士一定会被传送门华丽丽地玩死了；

3.剪枝。这里的剪枝因为两层之间传送不消耗时间，所以我们仅仅须要求：T(当前位置)+DIS(公主位置-当前位置)<=T才干继续进行，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;
}