poj1682

题目大意:在三个河岸X,Y,Z上分别有n,m,k个部落,要求用最少的费用,使每个部落都能通过至少一座桥与其它河岸联系.

     期中桥梁不能交叉.每条桥梁需要的费用为两个部落海拔差的绝对值.

思路：两两进行dp。。然后枚举交点在进行dp一次。。

         即多重dp

/*
  State:Accepted
  Time:2013-03-08 20:49:50
*/
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#define MXN  105
using namespace std;
int a[MXN],b[MXN],f[MXN][MXN] , n , m ,p ,ans , T ;

int max(int x , int y){ 
    return x > y? x : y;
};

void init(){
      scanf("%d%d",&n, &m);
      for (int i = 1; i <= n; ++i)  
          scanf("%d" , &a[i]);   
      for (int i = 1; i <= m; ++i)  
          scanf("%d" , &b[i]);   
}

void dp(){
      memset(f , 0 ,sizeof(f));
      for (int i = 2; i <= n; ++i)
         for (int j = 2; j <= m; ++j){
             f[i][j] = max(f[i - 1][j] , f[i][j - 1]);
             f[i][j] = max(f[i][j] , f[i - 1][j -1]);
             if (a[i]!=b[j]){
                   int k1 , k2;
                   for (k1 = i - 1; k1 > 0 ; --k1)
                       if (a[k1] == b[j]) break;
                   for (k2 = j - 1; k2 > 0; --k2)
                       if (a[i] == b[k2]) break;
                       
                   if (k1&&k2) f[i][j] = max(f[i][j] , f[k1 - 1][k2 - 1] + 2);
                     
            }
         }

      printf("%d\n",f[n][m]);
}

int main(){
      freopen("poj1692.in","r",stdin);
      freopen("poj1692.out","w",stdout);
      scanf("%d",&T);
      for (int i = 1; i <= T; ++i){
          init();
          dp();
      }
      fclose(stdin); fclose(stdout);
}