HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化)

HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化)

题意分析

给出一系列的石头的数量，然后问石头能否被平分成为价值相等的2份。首先可以确定的是如果石头的价值总和为奇数的话，那么肯定不能被平分。若为偶数，则对valuesum/2为背包容量，全体石头为商品做完全背包。把完全背包进行二进制优化后，转为01背包即可。

代码总览

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define nmax 20005 * 6
#define INIT(x,y) memset(x,y,sizeof(x))
using namespace std;
int c[6];
int val[150],num[150];
int dp[nmax];
int main()
{
    //freopen("in.txt","r",stdin);
    int cas = 0;
    while(1){
        INIT(val,0);INIT(num,0);INIT(c,0);
        int judge = false;
        int sum = 0;
        for(int i = 0; i<6;++i) {scanf("%d",&c[i]); sum+=c[i] * (i+1);}
        if(sum == 0) break;
        printf("Collection #%d:
",++cas);
        int cnt = 0;
        if(sum %2 == 0){
            for(int i = 0 ;i<6;++i){
                for(int j =1; j<=c[i]; j<<=1){
                    val[cnt] = j*(i+1);
                    num[cnt++] = j;
                    c[i]-=j;
                }
                if(c[i]>0){
                    val[cnt] = c[i] * (i+1);
                    num[cnt++] = c[i];
                }
            }
            INIT(dp,0);
            dp[0] = 1;
            for(int i = 0; i<=cnt ;++i){
                for(int j = sum; j>=val[i];--j)
                    if(dp[j-val[i]]) dp[j] = 1;
            }
            if(dp[sum/2] == 1) judge = true;
        }
        if(judge) printf("Can be divided.

");
        else printf("Can't be divided.

");

    }
    return 0;
}