ＴＣSRM 593 div2（１０００）（ｄｐ）

Problem Statement
	The pony Rainbow Dash wants to choose her pet. There are N animals who want to be her pet. Rainbow Dash numbered them 0 through N-1. To help her make the decision, Rainbow Dash decided to organize a relay race for the animals. The race track is already known, and for each animal we know how fast it is. More precisely, you are given vector <int>s A and B with the following meaning: For each i, the animal number i will take between A[i] and B[i] seconds (inclusive) to complete the track. For the race the animals will be divided into two competing teams. This is a relay race, so the team members of each team will all run the same track, one after another -- when the first team member finishes, the second one may start, and so on. Thus the total time in which a team completes the race is the sum of the times of all team members. Note that we can use the estimates given by A and B to estimate the total time for any team of animals. Given two teams S and T, the value maxdiff(S,T) is defined as the largest possible difference in seconds between the time in which team S finishes the course and the time in which team T finishes the course. Rainbow Dash now needs to assign each of the animals to one of the two competing teams. She wants to see a close competition, so she wants the teams to finish as close to each other as possible. Formally, she wants to divide all animals into teams S and T in a way that minimizes maxdiff(S,T). Return the smallest possible value of maxdiff(S,T).
Definition
	Class: MayTheBestPetWin Method: calc Parameters: vector <int>, vector <int> Returns: int Method signature: int calc(vector <int> A, vector <int> B) (be sure your method is public)
Notes
-	The teams are not required to contain the same number of animals.
Constraints
-	A will contain between 2 and 50 elements, inclusive.
-	A and B will contain the same number of elements.
-	Each element of A will be between 1 and 10,000, inclusive.
-	Each element of B will be between 1 and 10,000, inclusive.
-	For each i, B[i] will be greater than or equal to A[i].
Examples
0)
	{3,4,4,7} {3,4,4,7} Returns: 2 In this test case we know the exact time in which each of the animals completes the track. An optimal solution is to choose teams S={0,3} and T={1,2}. Then team S will certainly complete the track in 3+7 = 10 seconds, and team T in 4+4 = 8 seconds. Thus, maxdiff(S,T)=2.
1)
	{1,3,5,4,5} {2,5,6,8,7} Returns: 5 Here one of the optimal solutions is S={2,3} and T={0,1,4}. For these two teams we have maxdiff(S,T)=5. For example, it is possible that S will complete the track in 6+8 = 14 seconds, and T will complete it in 1+3+5 = 9 seconds. It is also possible that S will complete the track up to 5 seconds before T does.
2)
	{2907,949,1674,6092,8608,5186,2630,970,1050,2415,1923,2697,5571,6941,8065,4710,716,756,5185,1341,993,5092,248,1895,4223,1783,3844,3531,2431,1755,2837,4015} {7296,6954,4407,9724,8645,8065,9323,8433,1352,9618,6487,7309,9297,8999,9960,5653,4721,7623,6017,7320,3513,6642,6359,3145,7233,5077,6457,3605,2911,4679,5381,6574} Returns: 52873

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木想出来 

按和来进行背包 这样来看吧 如果按a数组来背 那我们要算出这几个值

s1(前面背出来的和值）    s2（剩余的） a数组

s3（前面背出来的和值）   s4（剩余的） b数组

想求s3-s2  和s1-s4 最值也就从这两种情况里取一个 

如果按和背的话 直接就保存了 s1+s3 而（s2+s1)和(s3+s4)是定值 每次背的值 s1+s3-(s2+s1) 这样就出了s3-s2 同样s1-s4也求出来了 

so easy...

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 using namespace std;

 8 #define INF 0xfffffff

 9 int dp[1000010];

10 class MayTheBestPetWin

11 {

12     public:

13     int calc(vector <int> A, vector <int> B)

14     {

15         int n = A.size(),s1=0,s2=0,i,j;

16         for(i = 0; i < n ; i++)

17         {

18             s1+=A[i];

19             s2+=B[i];

20             A[i]+=B[i];

21         }

22         int v = s1+s2;

23         memset(dp,0,sizeof(dp));

24         dp[0] = 1;

25         for(i = 0; i < n ; i++)

26             for(j = v ; j>=A[i] ; j--)

27             dp[j]=max(dp[j],dp[j-A[i]]);

28         int ans = INF;

29         for(i = 1 ; i <= v ; i++)

30         {

31             if(dp[i]>0)

32             {

33                 ans = min(ans,max(abs(i-s1),abs(i-s2)));

34             }

35         }

36         return ans;

37     }

38 };

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