HDU 3533 Escape（BFS+模擬）兩種思路

Escape

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2400    Accepted Submission(s): 693

Problem Description

The students of the HEU are maneuvering for their military training.
The red army and the blue army are at war today. The blue army finds that Little A is the spy of the red army, so Little A has to escape from the headquarters of the blue army to that of the red army. The battle field is a rectangle of size m*n, and the headquarters of the blue army and the red army are placed at (0, 0) and (m, n), respectively, which means that Little A will go from (0, 0) to (m, n). The picture below denotes the shape of the battle field and the notation of directions that we will use later.



The blue army is eager to revenge, so it tries its best to kill Little A during his escape. The blue army places many castles, which will shoot to a fixed direction periodically. It costs Little A one unit of energy per second, whether he moves or not. If he uses up all his energy or gets shot at sometime, then he fails. Little A can move north, south, east or west, one unit per second. Note he may stay at times in order not to be shot.
To simplify the problem, let’s assume that Little A cannot stop in the middle of a second. He will neither get shot nor block the bullet during his move, which means that a bullet can only kill Little A at positions with integer coordinates. Consider the example below. The bullet moves from (0, 3) to (0, 0) at the speed of 3 units per second, and Little A moves from (0, 0) to (0, 1) at the speed of 1 unit per second. Then Little A is not killed. But if the bullet moves 2 units per second in the above example, Little A will be killed at (0, 1).
Now, please tell Little A whether he can escape.

 

Input

For every test case, the first line has four integers, m, n, k and d (2<=m, n<=100, 0<=k<=100, m+ n<=d<=1000). m and n are the size of the battle ground, k is the number of castles and d is the units of energy Little A initially has. The next k lines describe the castles each. Each line contains a character c and four integers, t, v, x and y. Here c is ‘N’, ‘S’, ‘E’ or ‘W’ giving the direction to which the castle shoots, t is the period, v is the velocity of the bullets shot (i.e. units passed per second), and (x, y) is the location of the castle. Here we suppose that if a castle is shot by other castles, it will block others’ shots but will NOT be destroyed. And two bullets will pass each other without affecting their directions and velocities.
All castles begin to shoot when Little A starts to escape.
Proceed to the end of file.

 

Output

If Little A can escape, print the minimum time required in seconds on a single line. Otherwise print “Bad luck!” without quotes.

 

Sample Input

4 4 3 10N 1 1 1 1W 1 1 3 2W 2 1 2 44 4 3 10N 1 1 1 1W 1 1 3 2W 1 1 2 4

 

Sample Output

9Bad luck!

 

中文題意：

    

一個人從（0,0）跑到(n,m),只有k點能量，一秒消耗一點，在圖中有k個炮塔，給出炮塔的射擊方向c，射擊間隔t，子彈速度v，座標x,y

問這個人能不能安全到達終點

要求: 

1.人不能到達炮塔所在的座標

2.炮塔會擋住子彈

3.途中遇到子彈是安全的，但是人如果停在這個座標，而子彈也剛好到這個座標，人就被射死

4.人可以選擇停止不動

思路：其實不難，我們只需要看當人位於某個點的時候，其四個方向是否有炮塔，這個炮塔是都向人的方向射擊，然後再看子彈是否剛好位於這個座標即可。

而標記的話，vis[x][y][time]，對於time時刻，人位於x,y的情況只需要訪問一次，這是唯一的

兩種思路：

    第一種：根據自己的位置，向四個方向找最近的堡壘，然後判斷是否會被射殺

    第二種：遍歷全部堡壘數，看子彈是否會被阻擋，在看是否能夠射擊到。（這種比較快，爲何？？？）

第一種

#include<bits/stdc++.h>

using namespace std;
int m,n,k,d;
bool maps[110][110];			//堡壘標記 
bool vis[110][110][1010];		//訪問的點座標還有時間【x】【y】【time】 

int dir[5][2]={{1,0},{-1,0},{0,1},{0,-1},{0,0}};  //上下左右還有停止不動 
struct node{
	char d;		//方向 
	int t,v;	//發射週期，子彈速度，堡壘座標 
}castle[110][110];

struct node1{	 
	int x,y;
	int step;
};

bool judge1(int x,int y,int time){
	//向上,堡壘向下射擊,並且判斷是否會被射中 
	for(int i=x-1;i>=0;i--){
		if(maps[i][y]){
			if(castle[i][y].d=='S' && time>=1.0*(x-i)/castle[i][y].v){
				double c=1.0*(x-i)/castle[i][y].v;
				if(time==c) return false;
				while(time>=c){
					if(time==c) return false;
					c+=castle[i][y].t;
				}
			}
			break;					//距離最近的子彈纔不會被別的堡壘擋道 
		}
	} 
	//向下，堡壘向上射擊 
	for(int i=x+1;i<=m;i++){
		if(maps[i][y]){
		  	if(castle[i][y].d=='N' && time>=1.0*(i-x)/castle[i][y].v){
				double c=1.0*(i-x)/castle[i][y].v;
				if(time==c) return false;
				while(time>=c){
					if(time==c) return false;
					c+=castle[i][y].t;
				}
			}
			break;
		}
	}
	//向左，堡壘向右射擊
	for(int i=y-1;i>=0;i--){
		if(maps[x][i]){
			if(castle[x][i].d=='E' && time>=1.0*(y-i)/castle[x][i].v){
				double c=1.0*(y-i)/castle[x][i].v;
				if(time==c) return false;
				while(time>=c){
					if(time==c) return false;
					c+=castle[x][i].t;
				}
			}
			break;
		}
	} 
	//向右，堡壘向左射擊
	for(int i=y+1;i<=n;i++){
		if(maps[x][i]){
			if(castle[x][i].d=='W' && time>=1.0*(i-y)/castle[x][i].v){
				double c=1.0*(i-y)/castle[x][i].v;
				if(time==c) return false;
				while(time>=c){
					if(time==c) return false;
					c+=castle[x][i].t;
				}
			}
			break;
		}
	}
	return true;
}

//判斷這個時候會不會跟子彈撞上 
bool judge(int x,int y,int time){
	if(x<0 || x>m || y<0 || y>n || time>d || maps[x][y]){		//在規定時間內要逃出（time<d） 
		return false;
	}
	if(judge1(x,y,time)){
		
		return true;
	}
	return false;
}
//BFS模板 
void BFS(){
	memset(vis,0,sizeof vis);
	
	queue<node1> s;
	s.push(node1{0,0,0});
	vis[0][0][0]=1;
	while(!s.empty()){
		node1 cnt = s.front();
		s.pop();
		if(cnt.step>d) printf("Bad luck!\n");
		if(cnt.x==m && cnt.y==n && cnt.step<=d){
			printf("%d\n",cnt.step);
			return;
		}
		for(int i=0;i<5;i++){
			int x=cnt.x+dir[i][0];
			int y=cnt.y+dir[i][1];
			int time=cnt.step+1;
			if(judge(x,y,time) && !vis[x][y][time]){
				vis[x][y][time]=1;
				s.push(node1{x,y,time});
			}
		}
	}
	printf("Bad luck!\n");
}

int main(){
	char a;
	int b,c,e,f;
	while(~scanf("%d%d%d%d",&m,&n,&k,&d)){
		memset(maps,0,sizeof maps);
		for(int i=0;i<k;i++){
			getchar();
			scanf("%c%d%d%d%d",&a,&b,&c,&e,&f);
			castle[e][f].d=a,castle[e][f].t=b,castle[e][f].v=c;
			maps[e][f]=1;
		}
		BFS();
	}
} 

第二種：

#include<bits/stdc++.h>

using namespace std;
int m,n,k,d;
bool maps[110][110];
bool vis[110][110][1010];

int dir[5][2]={{1,0},{-1,0},{0,1},{0,-1},{0,0}};
struct node{
	char d;
	int t,v,x,y;	//發射週期，子彈速度，堡壘座標 
}castle[110];

struct node1{
	int x,y;
	int step;
};

bool judge1(int x,int y,int time){
	for(int i=0;i<k;i++){
		//同一x軸上，方向向W的射擊 
		if(x==castle[i].x && castle[i].y>y && castle[i].d=='W' &&time>=1.0*(castle[i].y-y)/castle[i].v){
			//查看是否會被擋住 
			int f=0;
			for(int j=castle[i].y-1;j>y;j--){
				if(maps[x][j]){
					f=1;
					break;
				}
			}
			if(f) continue;
			float c=1.0*(castle[i].y-y)/castle[i].v;
			if(c==time) return false;
			while(time>=c){
				if(c==time) return false;
				c+=castle[i].t;
			}
			continue;
		}
		//同x軸，方向向E 
		if(x==castle[i].x && castle[i].y<y && castle[i].d=='E' && time>=1.0*(y-castle[i].y)/castle[i].v){
			int f=0;
			for(int j=castle[i].y+1;j<y;j++){
				if(maps[x][j]){
					f=1;
					break;
				}
			}
			if(f) continue;
			double c=1.0*(y-castle[i].y)/castle[i].v;
			if(c==time) return false;
			while(time>=c){
				if(c==time) return false;
				c+=castle[i].t;
			}
			continue;
		}
		if(y==castle[i].y && castle[i].x<x && castle[i].d=='S' && time>=1.0*(x-castle[i].x)/castle[i].v){
			int f=0;
			for(int j=castle[i].x+1;j<x;j++){
				if(maps[j][y]){
					f=1;
					break; 
				}
			}
			if(f) continue;
			double c=1.0*(x-castle[i].x)/castle[i].v;
			if(c==time) return false;
			while(time>=c){
				if(c==time) return false;
				c+=castle[i].t;
			}
			continue;
		}
		if(y==castle[i].y && castle[i].x>x && castle[i].d=='N' && time>=1.0*(castle[i].x-x)/castle[i].v){
			int f=0;
			for(int j=castle[i].x-1;j>x;j--){
				if(maps[j][y]){
					f=1;
					break;
				}
			}
			if(f) continue;
			double c=1.0*(castle[i].x-x)/castle[i].v;
			if(c==time) return false;
			while(time>=c){
				if(c==time) return false;
				c+=castle[i].t;
			}
			continue;
		}
	}
	return true;
}

//判斷這個時候會不會跟子彈撞上 
bool judge(int x,int y,int time){
	if(x<0 || x>m || y<0 || y>n || time>d || maps[x][y]){
		return false;
	}
	if(judge1(x,y,time)){
		return true;
	}
	return false;
}

void BFS(){
	memset(vis,0,sizeof vis);
	queue<node1> s;
	s.push(node1{0,0,0});
	vis[0][0][0]=1;
	while(!s.empty()){
		node1 cnt = s.front();
		s.pop();
		if(cnt.x==m && cnt.y==n && cnt.step<=d){
			printf("%d\n",cnt.step);
			return;
		}
		if(cnt.step>d) printf("Bad luck!\n");
		for(int i=0;i<5;i++){
			int x=cnt.x+dir[i][0];
			int y=cnt.y+dir[i][1];
			int time=cnt.step+1;
            if(!vis[x][y][time]){
				if(judge(x,y,time)){
					vis[x][y][time]=1;
					s.push(node1{x,y,time});
				}
			}
		}
	}
	printf("Bad luck!\n");
}

int main(){
	while(~scanf("%d%d%d%d",&m,&n,&k,&d)){
		memset(maps,0,sizeof maps);
		for(int i=0;i<k;i++){
			getchar();
			scanf("%c%d%d%d%d",&castle[i].d,&castle[i].t,&castle[i].v,&castle[i].x,&castle[i].y);
			maps[castle[i].x][castle[i].y]=1;
		}
		BFS();
	}
}