Lightoj1031_ interval DP

Title Link: http://lightoj.com/volume_showproblem.php?problem=1031

Title Description:

Give a series, two people take turns to fetch the number, finish. Can take a lot of numbers at a time, but only from the first or the end of the beginning to take a number of consecutive. Q. What is the maximum value of the last two?

Interval DP;

I've been reading a few ladder reports before I figured it out, and now I think it fits the requirements of dynamic planning.

D (I, j) represents the maximum value that a number of people can fetch in the array I to J, and then enumerates the split points in the middle,

ans = max (ans, sum[k]-sum[i-1]-d (k+1, J));
ans = max (ans, sum[j]-sum[k-1]-d (i, k-1));

Ways to adopt and memorize searches

1#include <algorithm>2#include <iostream>3#include <cstring>4#include <cstdlib>5#include <cstdio>6#include <vector>7#include <ctime>8#include <queue>9#include <list>Ten#include <Set> One#include <map> A using namespacestd; - #defineINF 0x3f3f3f3f -typedefLong LongLL; the  - inta[ the], sum[ the], dp[ the][ the]; - intSolveintLintR) - { +     if(Dp[l][r]! =-1*INF) -         returnDp[l][r]; +     intres = sum[r]-sum[l-1]; A      for(inti = l; I <= R; i++) at     { -res = max (res, sum[i]-sum[l-1]-solve (i+1, R)); -res = max (res, sum[r]-sum[i-1]-solve (L, I-1)); -     } -DP[L][R] =Res; -     returnRes; in } - intMain () to { +     intt, N; -scanf"%d", &t); the      for(intCA =1; CA <= T; ca++) *     { $scanf"%d", &n);Panax Notoginsengsum[0] =0; -          for(inti =1; I <= N; i++) the              for(intj = i; J <= N; J + +) +DP[I][J] =-1*INF; A          for(inti =1; I <= N; i++) the         { +scanf"%d", &a[i]); -Dp[i][i] =A[i]; $Sum[i] = sum[i-1] +A[i]; $         } -printf"Case %d:%d\n", CA, Solve (1, N)); -     } the     return 0; -}

View Code

Lightoj1031_ interval DP