HDU 1059（多重背包）

题意描述

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

思路

一道多重背包的问题，先计算所有的和，如果不能被平分，直接输出。否则就求dp[sum/2]是否等于sum/2，如果相等就说明可以平分。

AC代码

#include<iostream>
#include<string>
#include<cstring>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=15;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
int a[7],dp[M],v[M];
int Case=1;
void solve(){
    while(cin>>a[1]){
        cin>>a[2]>>a[3]>>a[4]>>a[5]>>a[6];
        mst(dp,0);
        int sum=0;
        rep(i,1,7) sum+=a[i]*i;
        if(sum==0) break;
        if(sum%2!=0) printf("Collection #%d:nCan't be divided.nn",Case++);
        else{
            int cnt=1;
            rep(i,1,7){
                int k=1;
                while(k<=a[i]){
                    v[cnt++]=k*i;
                    a[i]-=k;
                    k*=2;
                }
                if(a[i]>0){
                    v[cnt++]=i*a[i];
                }
            }
            int n=cnt;
            rep(i,1,n){
                rrep(j,sum/2,v[i]){
                    dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
                }
            }
            if(dp[sum/2]!=sum/2) printf("Collection #%d:nCan't be divided.nn",Case++);
            else printf("Collection #%d:nCan be divided.nn",Case++);
        }

    }
}
int main(){
    //IOS;
    solve();
    return 0;
}