UVA 10944 Nuts for nuts..

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=21&page=show_problem&problem=1885

Dynamic Programming. dp[i][j]表示以ith nut为结尾，状态j下的最少步数。假设有n个nuts，状态有(2^(n-1))-1个。一步一步的添加得到最后结果。代码如下：

 #include <iostream>
 #include <math.h>
 #include <stdio.h>
 #include <cstdio>
 #include <algorithm>
 #include <string.h>
 #include <string>
 #include <sstream>
 #include <cstring>
 #include <queue>
 #include <vector>
 #include <functional>
 #include <cmath>
 #include <set>
 #define SCF(a) scanf("%d", &a)
 #define IN(a) cin>>a
 #define FOR(i, a, b) for(int i=a;i<b;i++)
 #define Infinity 999999999
 typedef long long Int;
 using namespace std;

 struct point {
     int x, y;
 };

 int x, y;
 point nuts[];
 int num = ;
 int dp[][];
 int dis[][];

 int main()
 {
     while (scanf("%d %d", &x, &y) != EOF)
     {
         num = ;
         getchar();
         FOR(i, , x)
         {
             FOR(j, , y)
             {
                 char c = getchar();
                 if (c == 'L')
                 {
                     nuts[].x = i;
                     nuts[].y = j;
                 }
                 else if (c == '#')
                 {
                     nuts[num].x = i;
                     nuts[num++].y = j;
                 }
             }
             getchar();
         }

         if (num == )
         {
             printf("%d\n", );
             continue;
         }

         FOR(i, , num)
         {
             FOR(j, i, num)
             {
                 dis[i][j] = dis[j][i] = max(abs(nuts[i].x - nuts[j].x), abs(nuts[i].y - nuts[j].y));
             }
         }

         int states = ( << (num - )) - ;
         for (int s = ; s <= states; s++)
         {
             for (int i = ; i < num; i++)
             {
                 dp[i][s] = Infinity;
             }
         }

         for (int i = ; i < num; i++)
             dp[i][ << (i - )] = dis[][i];

         for (int i = ; i <= states; i++)
         {
             for (int j = ; j < num; j++)
             {
                 if (i & ( << (j - )))
                 {
                     for (int k = ; k < num; k++)
                     {
                         if (!(i & ( << (k - ))))
                         {
                             if (dp[k][i + ( << (k - ))] > dp[j][i] + dis[j][k])
                                 dp[k][i + ( << (k - ))] = dp[j][i] + dis[j][k];
                         }
                     }
                 }
             }
         }
         int finalAns = Infinity;
         for (int i = ; i < num; i++)
         {
             if (dp[i][states] + dis[i][] < finalAns)
                 finalAns = dp[i][states] + dis[i][];
         }
         printf("%d\n", finalAns);

     }
     return ;
 }