QuXiao

    地圖上有一些beeper，讓你從起點(diǎn)搜集所有beeper，再回到起點(diǎn)。但你只可以水平或者豎直的走，不能走對(duì)角線。讓你找出這樣走的最短路徑長度。
    因?yàn)榈貓D最大只有20×20，而beeper最多也只有10個(gè)，所以可以考慮用深搜，找到所有可能路徑，在其中加一些簡單的減枝就可以了。起初以為會(huì)超時(shí)，可最后還是0ms。：）

Source Code

Problem: 2907		User: QuXiao
Memory: 176K		Time: 0MS
Language: C++		Result: Accepted

• Source Code

#include <iostream>
#include <climits>
using namespace std;

struct Point
{
	int x, y;
};

int num, X, Y;
Point start, beeper[15];
int shortest;
int visited[15];


int Length (Point p1, Point p2)
{
	return abs(p1.x - p2.x) + abs(p1.y - p2.y);
}

void Input ()
{
	int i;
	cin>>X>>Y;
	cin>>start.x>>start.y;
	cin>>num;
	for (i=0; i<num; i++)
		cin>>beeper[i].x>>beeper[i].y;
}

void DFS (int cur, int len, int n)
{
	if ( n == num )
	{
		int t = Length(beeper[cur], start);
		if ( len +  t < shortest )
			shortest = len + t;
	}
	else if ( len < shortest )
	{
		int i;
		for (i=0; i<num; i++)
		{
			if ( visited[i] == 0 )
			{
				visited[i] = 1;
				DFS (i, len+Length(beeper[cur], beeper[i]), n+1);
				visited[i] = 0;
			}
		}
	}
}


void Solve ()
{
	int i, t;
	shortest = INT_MAX;
	memset(visited, 0, sizeof(visited));
	for (i=0; i<num; i++)
	{
		t = Length(beeper[i], start);
		visited[i] = 1;
		DFS (i, t, 1);
		visited[i] = 0;
	}
	cout<<"The shortest path has length "<<shortest<<endl;
}

int main ()
{
	int test;
	cin>>test;
	while ( test-- )
	{
		Input ();
		Solve ();
	}

	return 0;
}