北大2907题

来源：互联网发布：网络理财骗局编辑：程序博客网时间：2024/05/18 08:43

题目连接：http://acm.pku.edu.cn/JudgeOnline/problem?id=2907

走点的先后次序会影响走的最短路径，点的个数最多10，因此用排列树穷举所有情况找到最短的找到各个点的距离。

#include <iostream>
#include <cmath>

using namespace std;

#define MAX 10

struct Position
{
int row;
int col;
};

struct Prob
{
int num;
int x[MAX];
Position pos[MAX];
Position org;

int len;
int res;

void Init()
{
len = 0;
res = 100000;

  int i;
  for(i = 0;i < MAX;++i)
   x[i] = i;
}

int Distance(Position& p1,Position& p2)
{
return abs(p1.col - p2.col) + abs(p1.row - p2.row);
}

void Work(int cur)
{
  if(cur == num)
  {
   if(len + Distance(pos[x[cur-1]],org) < res)
    res = len + Distance(pos[x[cur-1]],org);
  }
  else
  {
   int i,back;
   for(i = cur;i < num;++i)
   {
    back = len;
    swap(x[cur],x[i]);

    if(cur == 0)
     len = Distance(pos[x[cur]],org);
    else
     len += Distance(pos[x[cur-1]],pos[x[cur]]);
    if(len < res)
     Work(cur+1);

    len = back;
    swap(x[cur],x[i]);
   }
  }
}
};

int main()
{
freopen("in.txt","r",stdin);

Prob pb;
int i,t;
cin >> t;
while(t--)
{
  cin >> pb.len >> pb.len;
  cin >> pb.org.row >> pb.org.col;
  cin >> pb.num;

for(i = 0;i < pb.num;++i)
cin >> pb.pos[i].row >> pb.pos[i].col;

pb.Init();
pb.Work(0);

printf("The shortest path has length %d/n",pb.res);
}
return 0;
}