TOJ-1356 Tug of War

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.

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.

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

Output for Sample Input

190 200

Source: Waterloo Local Contest Sep. 30, 2000

这道题我们看作一个背包的容量是总质量的一半，并且要放总数量的一半的背包问题。

 

dp[j][k]=1状态表示背包放入k个物品时总质量为j。

#include <stdio.h>
#include <iostream>
#include <memory.h> 
using namespace std;
int p[110],dp[45010][110];
int main(){
    int i,j,k,n,sum,m;
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        sum = 0;
        for(i=1;i<=n;i++){
            scanf("%d",&p[i]);
            sum += p[i];
        }
        m = (n+1)/2;
        dp[0][0] = 1;
        for(i=1;i<=n;i++){
            for(j=sum/2;j>=p[i];j--){
                for(k=m;k>0;k--){
                    if(dp[j-p[i]][k-1])
                        dp[j][k] = 1;
                }
            }
        }
        for(i=sum/2;i>0;i--){
            if(n%2==0)  
            {  
                if(dp[i][m]==1)  
                    break;  
            }  
            else  
            {  
                if(dp[i][m]==1||dp[i][m-1]==1)  
                    break;  
            }
        }
        printf("%d %d\n",i,sum-i); 
    }
    return 0;
}

 

然而这道题的数据还是有漏洞的。由于限定两个背包中数量相差不超过1，所以将上述状态降为成如下状态是不行的：

dp[j]=k表示当背包内放k个物品时质量为j，该状态代码如下

#include <stdio.h>
#include <iostream>
#include <memory.h> 
using namespace std;
int p[110],dp[45010];
int main(){
    int i,j,k,n,sum,m;
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        sum = 0;
        for(i=1;i<=n;i++){
            scanf("%d",&p[i]);
            sum += p[i];
        }
        m = (n+1)/2;
        dp[0] = 1;
        for(i=1;i<=n;i++){
            for(j=sum/2;j>=0;j--){
                if(dp[j] && dp[j]<=m && j+p[i]<=sum/2){  
                    if(!dp[j+p[i]])  
                        dp[j+p[i]] = dp[j]+1;  
                    else  
                        dp[j+p[i]] = min(dp[j+p[i]],dp[j]+1);  
                }
            }
        }
        for(i = sum/2;i>=0;i--)  
            if(dp[i]){  
            break;  
        } 
        printf("%d %d\n",i,sum-i); 
    }
    return 0;
}

显然这个代码无法限制两背包数量之差。如4个物品质量为100，20，30，50时，输出应为80，120而不是100，100。