中文题面,题意略
坑点:两个传送门之间不能传,WA到天荒地老~~~~
代码:
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define ll long long
#define mod 10000000007
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
typedef pair <int,int> pii;
const int maxn = 1000+5 , inf = 0x3f3f3f3f;
char G[2][maxn][maxn];
bool vis[2][maxn][maxn];
int n,m,t;
const int dx[] = {1,-1,0,0};
const int dy[] = {0,0,-1,1};
bool can(int x,int y,int pos,int d){
if(1<=x&&x<=n&&1<=y&&y<=m&&!vis[pos][x][y]&&d<=t&&G[pos][x][y]!='*')
return true;
return false;
}
struct Node{
int x,y,d,pos;
Node(int xx,int yy,int dd,int ee):x(xx),y(yy),d(dd),pos(ee){};
};
void bfs(){
queue<Node>q;
q.push(Node(1,1,0,0));
vis[0][1][1] = 1;
while(!q.empty()){
Node now = q.front();q.pop();
if(G[now.pos][now.x][now.y]=='P'){
cout<<"YES"<<endl;
return;
}
for(int i=0;i<4;i++){
int nx = now.x+dx[i];
int ny = now.y+dy[i];
if(G[now.pos][nx][ny]=='#'){
if(G[!now.pos][nx][ny]!='*'&&G[!now.pos][nx][ny]!='#'&&can(nx,ny,now.pos,now.d+1)&&!vis[!now.pos][nx][ny]){
vis[!now.pos][nx][ny] = 1;
vis[now.pos][nx][ny] = 1;
q.push(Node(nx,ny,now.d+1,!now.pos));
}
}else{
if(can(nx,ny,now.pos,now.d+1)){
vis[now.pos][nx][ny] = 1;
q.push(Node(nx,ny,now.d+1,now.pos));
}
}
}
}
cout<<"NO"<<endl;
}
int main(){
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
while(T--){
mem(G);mem(vis);
scanf("%d%d%d",&n,&m,&t);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++){
scanf(" %c",&G[0][i][j]);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++){
scanf(" %c",&G[1][i][j]);
}
bfs();
}
}