HDU 1253 - 胜利大逃亡

Problem: a* b* c 立方体，从（1,1,1）到（a,b,c），最短路<=限制时间

Solution: BFS

Code:

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline ll read() {
	ll x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9') {
		if(ch=='-') f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9') {
		x=(x<<1)+(x<<3)+ch-'0';
		ch=getchar();
	}
	return x*f;
}
int t,a,b,c,ti,ans;
int mp[55][55][55];
int vis[55][55][55];
int dx[6]={-1,1,0,0,0,0};
int dy[6]={0,0,-1,1,0,0};
int dz[6]={0,0,0,0,-1,1};
struct node{
	int x,y,z,cnt;
};
int solve(int x,int y,int z){
	node p={x,y,z,0};
	queue<node>q;
	q.push(p);
	vis[x][y][z]=1;
	while(!q.empty()){
		node t=q.front();
		q.pop();
		if(t.x==a&&t.y==b&&t.z==c&&t.cnt<=ti) return t.cnt;
		for(int i=0;i<6;i++){
			int xx=t.x+dx[i],yy=t.y+dy[i],zz=t.z+dz[i];
			if(xx<1||xx>a||yy<1||yy>b||zz<1||zz>c||vis[xx][yy][zz]||mp[xx][yy][zz]) continue;
		    vis[xx][yy][zz]=1;
		    q.push({xx,yy,zz,t.cnt+1});
		}
	}
	return -1;
}
int main() {
	t=read();
	while(t--){
		a=read(),b=read(),c=read(),ti=read();
		for(int i=1;i<=a;i++){
			for(int j=1;j<=b;j++){
				for(int k=1;k<=c;k++){
					mp[i][j][k]=read();
				}
			}
		} 
		memset(vis,0,sizeof(vis));
		ans=solve(1,1,1);
		printf("%d\n",ans);
	}
	return 0;
}

标签：HDU,ch,&&,read,逃亡,yy,xx,1253,zz
来源： https://blog.csdn.net/Frankie_Ku/article/details/115419470