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:\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;}