Tug of War
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 8500 | Accepted: 2311 |
Description
A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight
of the people on each team should be as nearly equal as possible.
Input
The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the
picnic.
Output
Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.
Sample Input
3 100 90 200
Sample Output
190 200
Source
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[45010][105],w[105];
int main()
{
int i,j,k,n,sum,sum1,sum2,num;
while(cin>>n)
{
sum=0;
for(i=1;i<=n;i++)
{
cin>>w[i];
sum+=w[i];
}
sum1=(sum+1)/2;
num=(n+1)/2;
memset(dp,0,sizeof(dp));
//cout<<"ok"<<endl;
for(i=1;i<=n;i++)
{
for(j=sum1;j>=w[i];j--)
{
for(k=num;k>=1;k--)
{
dp[j][k]=max(dp[j][k],dp[j-w[i]][k-1]+w[i]);
}
}
}
if(dp[sum1][num]>sum-dp[sum1][num])
{
cout<<sum-dp[sum1][num]<<" "<<dp[sum1][num]<<endl;
}
else
{
cout<<dp[sum1][num]<<" "<<sum-dp[sum1][num]<<endl;
}
}
}
不知道怎麼回事?用二維費用揹包竟然過了，應該過不了！
5 50 2 25 21 2 正確答案應該是48 52正確的應該是必須在n/2人之內找到重量不大於總重量一半的重量！
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[45010][105],w[105];
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int i,j,k,n,sum,sum1,sum2,num,ans;
while(cin>>n)
{
sum=0;
for(i=1;i<=n;i++)
{
cin>>w[i];
sum+=w[i];
}
sum1=(sum+1)/2;
num=(n+1)/2;
memset(dp,-1,sizeof(dp));
dp[0][0]=0;
//cout<<"ok"<<endl;
for(i=1;i<=n;i++)
{
for(j=sum1;j>=w[i];j--)
{
for(k=num;k>=1;k--)
{
if(dp[j-w[i]][k-1]!=-1)
dp[j][k]=max(dp[j][k],dp[j-w[i]][k-1]+w[i]);
}
}
}
ans=0;
for(i=0;i<=sum1;i++)
{
if(dp[i][num]>ans)
{
ans=dp[i][num];
}
if(dp[i][num-1]>ans)
{
ans=dp[i][num-1];
}
}
if(ans<sum-ans)
{
cout<<ans<<" "<<sum-ans<<endl;
}
else
{
cout<<sum-ans<<" "<<ans<<endl;
}
}
}
