Algorithm increases the maximum value

Problem DescriptionTo give n ordered integers to Ai bi, you need to select some integers to make all of the Ai+bi and max of the number you have selected. Also ask you to select the number of pairs of AI and non-negative, bi and non-negative.　　Input format input The first behavior n, the number of pairs of the following n lines per line two integer AI bi output format output you selected pairs of Ai+bi and sample input 5-403-625-847 901-624-708-293 413886 709 Sample output 1715 data size and conventions 1<=n<=100-1000<=ai,bi<=1000 01 Backpack Variant, copy the other people's ideas: not directly calculate the Ai+bi of the selected number of and, but convert to calculate in AI and under certain circumstances try to make the selected bi and Max. So it became a problem similar to the 01 backpack. First filter out all AI and bi are less than 0 pairs, so that dp[i][j] represents the first I number of pairs, the selected AI and for the case of J, the maximum value of BI, will dp[i][j] initialized to-inf, and then initialize all known legal conditions: dp[i][a[i]] = B[i], DP[I][J] = max (Dp[i-1][j], dp[i][j]), if j-a[i] exists, dp[i][j] = max (Dp[i][j], Dp[i-1][j-a[i]] + b[i]). To avoid a negative number as an array subscript, add an offset.
Source: http://www.cnblogs.com/wangyiming/p/6476019.html

1#include <stdio.h>2 3 #defineMax (A, B) ((A&GT;B)? ( A):(B))4 #defineN 1000005 6 voidInputint[] ,int[] ,int );7 intMaxnum (int[] ,int[] ,int );8 9 intMainvoid)Ten { One     intN; Ascanf"%d", &n); -      -     intNa[n], nb[n]; the input (Na, Nb, n); -      -     intAns =Maxnum (Na, Nb, n); -printf"%d\n", ans >=0? Ans:0 );  +     return 0;  - } +  A voidInputintNa[],intNb[],intN) { at      while(N-- ){ -scanf"%d%d", Na + +, Nb + + ); -     } - } -  - intMaxnum (intNa[],intNb[],intN) { in     Static intdp[ -][n*2+1] = {0} ; -     intres =-N; to     intI, J; +      for(i =0; I < n; i + + ){ -          for(j =-N; j <= N; j + +) ){ theDp[i][j + N] =-N; *         } $     }Panax Notoginseng      -      for(i =0; I < n; i + + ){ theDp[i][n + na[i]] =Nb[i]; +     } A      the      for(i =1; I < n; i + + ){ +          for(j =-N; j <= N; j + +) ){ -Dp[i][j + N] = Max (Dp[i][j + n], dp[i-1][j +N]); $              $             if(j + N-na[i] >=0&& J + N-na[i] < n2 ){ -Dp[i][j + N] = Max (Dp[i][j + n], dp[i-1][j + N-na[i]] +nb[i]); -             } the         } -     }Wuyi      the      for(j =0; J <= N; J + + ){ -         if(dp[n-1][j + N] >=0 ){ Wures = Max (res, j + dp[n-1][j +N]); -         } About     } $      -     returnRes; -}

Algorithm increases the maximum value