codevs1169, 51nod1084 (multi-threaded DP)

Let's just say codevs1169,

Title Link: http://codevs.cn/problem/1169/

Test instructions: Chinese question eh ~

Idea: Multi-threaded DP

Use Dp[i][j][k][l] to store a person in (I, j), a person at (K, L) position the maximum contribution to the answer, then the dynamic transfer equation is:

DP[I][J][K][L] = max (max (Dp[i-1][j][k-1][l], dp[i-1][j][k][l-1]), Max (Dp[i][j-1][k-1][l], dp[i][j-1][k][l- 1])) + A[i][j] + a[k][l]

Note: if (i, j) and (K, L) the same words can only a[i][j], a[k][l] only calculate one

Code:

1#include <iostream>2 using namespacestd;3 4 Const intMAXN =Wuyi;5 intA[MAXN][MAXN], DP[MAXN][MAXN][MAXN][MAXN];6 7 intMainvoid){8     intN, M;9CIN >> N >>m;Ten      for(inti =1; I <= N; i++){ One          for(intj =1; J <= M; J + +){ ACIN >>A[i][j]; -         } -     }    the      for(inti =1; I <= N; i++){ -          for(intj =1; J <= M; J + +){ -              for(intK =1; K <= N; k++){ -                  for(intL =1; L <= m; l++){ +DP[I][J][K][L] = max (max (Dp[i-1][j][k-1][L], Dp[i-1][j][k][l-1]), Max (Dp[i][j-1][k-1][L], Dp[i][j-1][k][l-1])); -                     if(i! = k | | J! = L) dp[i][j][k][l] + =A[k][l]; +Dp[i][j][k][l] + =A[i][j]; A                 } at             } -         } -     } -cout << Dp[n][m][n][m] <<Endl; -     return 0; -}

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51nod10084

Title Link: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1084

This problem is similar to the one above, but the data range is larger, with four-dimensional DP words, regardless of time complexity or space complexity are not allowed

It can be found that if you record the number of steps, you can count the number of steps and the number of rows, then it is similar to the above

Use dp[i][x1][x2] to record step I when the first person walked to line X1, the second person to go to the X2 when the answer to the greatest contribution

Then the dynamic transfer equation is:

DP[I][X1][X2] = max (max (dp[i-1][x1][x2], dp[i-1][x1][x2-1]), Max (dp[i-1][x1-1][x2], dp[i-1][x1-1][x2-1])) + a[x1][y1] + a[x2][y2]

where y1 = i-x1, y2 = i-x2

Note: (x1, y1) and (x2, y2) while A[x1][y1], A[x2][y2] can only calculate one

There is also a pit is the input n, M in the order of M, N//Pit me half a day

Code:

1#include <iostream>2 using namespacestd;3 4 Const intMAXN = 2e2 +Ten;5 intA[MAXN][MAXN], DP[MAXN <<1][MAXN][MAXN];6 7 intMainvoid){8     intN, M;9CIN >> M >>N;Ten      for(inti =1; I <= N; i++){ One          for(intj =1; J <= M; J + +){ ACIN >>A[i][j]; -         } -     } the      for(inti =1; I <= n + m; i++){ -          for(intX1 =1; X1 <= min (n, i); x1++){ -             intY1 = i-X1; -              for(intx2 =1; X2 <= min (n, i); x2++){ +                 inty2 = i-x2; -DP[I][X1][X2] = max (max (Dp[i-1][X1][X2], Dp[i-1][X1][X2-1]), Max (Dp[i-1][X1-1][X2], Dp[i-1][X1-1][X2-1])); +                 if(x1! = x2 | | y1! = y2) dp[i][x1][x2] + =A[x2][y2]; ADP[I][X1][X2] + =A[x1][y1]; at             } -         } -     } -cout << dp[n + m][n][n] <<Endl; -     return 0; -}

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codevs1169, 51nod1084 (multi-threaded DP)