poj2184 Cow Exhibition（带负数的背包问题）(关于背包问题的精髓可以去百度背包九讲)

Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7634		Accepted: 2775

Description

"Fat and docile, big and dumb, they look so stupid, they aren't much 
fun..." 
- Cows with Guns by Dana Lyons 

The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow. 

Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative. 

Input

* Line 1: A single integer N, the number of cows 

* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow. 

Output

* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0. 

Sample Input

5
-5 7
8 -6
6 -3
2 1
-8 -5

Sample Output

8

Hint

OUTPUT DETAILS: 

Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF 
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value 
of TS+TF to 10, but the new value of TF would be negative, so it is not 
allowed. 

要两边的和都大于零，并且求出最大值，其实就是背包问题，只不过带了负数，取100000处为原点就可

Memory: 1488 KB Time: 141 MS Language: G++ Result: Accepted

#include<stdio.h>

int c[1050],w[1050],f[201000];
#define A 1000000000
int main()
{
	int i,j,k,n,v,m,x,max;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<=200000;i++)
		{
			f[i]=-A;
		}//printf("%d ",w[0]);
		for(i=0;i<n;i++)
		{
			scanf("%d%d",c+i,w+i);
		}//printf("%d ",w[0]);
		
		f[100000]=0;
		for (i = 0; i < n; ++i)
		{
			if(c[i]<0&&w[i]<0)
				continue;
			if(c[i]>0)
			for (v = 200000; v >=c[i]; --v)//c[i]可优化为bound,bound = max {V - sum c[i,...n],c[i]}//正数的是从后往前的
			{
				if(f[v-c[i]]>-A)
				if(f[v] < f[v - c[i]] + w[i])
				{
					f[v] =f[v - c[i]] + w[i];
				}
			}
			else for (v = c[i];v<=200000+c[i]; v++)//注意啦，负数的递推方程是从前往后的，为的是避免重复加上一个状态
			{
				if(f[v-c[i]]>-A)
				if(f[v] < f[v - c[i]] + w[i])
				{
					f[v] =f[v - c[i]] + w[i];
				}
			}
			
		}
		max=-A;
		for(i=100000;i<=200000;i++)
		{
			if(f[i]>=0)
			{
				//printf("giuie");
				x=f[i]+i-100000;
				if(max<x)
					max=x;
			}
		}
		if(max!=-A)
			printf("%d\n",max);
		else printf("0\n");
	}
	return 0;
}

下面再贴一份学长写的16ms的程序，比我的快十倍啊

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<list>
#include<queue>
#define eps 1e-6
#define INF (1<<30)
#define PI acos(-1.0)
#define ll __int64
using namespace std;

int dp[210000]; //dp[i]表示当s的和为i-100000时，t的最大值
#define add 100000

int main()
{
   int n;

   while(scanf("%d",&n)!=EOF)
   {
      int s,t;
      int Max=0,Min=0;

      for(int i=0;i<=200000;i++)
         dp[i]=-INF;

      dp[add]=0;
      for(int i=1;i<=n;i++)
      {
         scanf("%d%d",&s,&t);
         if(s>0)//问题可能就出在这，他没有保存数组，而是输入一组数据就计算一下，可能时间就差在输入那个环节了，整体来看用的还是背包的思想
         {
            for(int i=Max+add;i>=Min+add;i--)
               dp[i+s]=max(dp[i+s],dp[i]+t);
            Max+=s;
         }
         else
         {
            for(int i=Min+add;i<=Max+add;i++)
               dp[i+s]=max(dp[i+s],dp[i]+t);
            Min+=s;
         }
      }
      int ans=0;

      for(int i=add;i<=Max+add;i++)
      {
         if(dp[i]>=0)
            ans=max(ans,i-add+dp[i]);
      }
      printf("%d\n",ans);


   }
   return 0;
}