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:\n",++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.\n\n");

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

    }

    return 0;

}