UVA 562(01背包)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=503

以sum/2为背包总量，结果为sum-d*dp[V]

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
using namespace std;

#define MEM(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define debug printf("!
")
#define INF 8000
#define MAX(a,b) a>b?a:b
#define blank pf("
")
#define LL long long
#define ep 1e-6

int dp[INF];

int way[INF][INF];//保存路径

int ci[INF];//容量
int wi[INF];//价值
int n,V,i,j,v,t,sum;
double G;


void zeroOnePack(int cost,int weight)
{
          for(v = V;v>=cost;v--)
          {
                    dp[v] =MAX(dp[v],dp[v-cost]+weight);
          }
}

int main()
{
          int t;
          sf("%d",&t);
          while(t--)
          {

                    MEM(dp,0);
                    MEM(way,0);
                    MEM(wi,0);

                    sf("%d",&n);

                    sum = 0;

                    for(i = 1;i<=n;i++)
                    {
                              sf("%d",&wi[i]);
                              sum+=wi[i];
                    }

                    G = (double)sum/2;
                    V = (int)G;

                    for(i = 1;i<=n;i++)
                    {
                              zeroOnePack(wi[i],wi[i]);
                    }

                    pf("sum:%d
",sum-2*dp[V]);

          }
    return 0;
}