Dividing-多重背包模板题

时间：2015-07-28 18:44:25 阅读：181 评论：0 收藏：0 [点我收藏+]

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Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Description

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.

Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0‘‘. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0‘‘; do not process this line.

Output

For each colletcion, output ``Collection #k:‘‘, where k is the number of the test case, and then either ``Can be divided.‘‘ or ``Can‘t be divided.‘‘.

Output a blank line after each test case.

Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

Sample Output

Collection #1:
Can‘t be divided.

Collection #2:
Can be divided.

原理，大家可以通过看背包九讲，我不认为自己比那大牛牛逼

所以我只能提供两种不同风格的代码给大家认识一下：

/*
Author: 2486
Memory: 1152 KB		Time: 0 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=20005;
int f[maxn<<2],v[maxn],w[maxn],n[maxn];
int a[6],mid;
void zeroonepack(int w,int v) {
    for(int i=mid; i>=w; i--) {
        f[i]=max(f[i],f[i-w]+v);
    }
}
void completepack(int w,int v) {
    for(int i=w; i<=mid; i++) {
        f[i]=max(f[i],f[i-w]+v);
    }
}
void multipack(int w,int v,int n) {
    if(w*n>mid) {
        completepack(w,v);
        return ;
    } else {
        for(int k=1; k<n; k<<=1) {
            zeroonepack(k*w,k*v);
            n-=k;
        }
        zeroonepack(n*w,n*v);
    }
}
int main() {
    int number=1;
    while(~scanf("%d%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5])) {
        int sum=0;
        for(int i=0; i<6; i++) {
            sum+=a[i]*(i+1);
        }
        mid=sum>>1;
        for(int i=0; i<=sum; i++) {
            f[i]=-maxn;
        }
        f[0]=0;
        if(!sum)break;
        printf("Collection #%d:\n",number++);
        if(sum&1) {
            printf("Can't be divided.\n");
        } else {
            for(int i=0; i<6; i++) {
                multipack(i+1,i+1,a[i]);
            }
            if(f[mid]==mid) {
                printf("Can be divided.\n");
            } else {
                printf("Can't be divided.\n");
            }
        }
        printf("\n");
    }
    return 0;
}

/*
Author: 2486
Memory: 3828 KB		Time: 47 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <string>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=200005;
int sizes[maxn<<2],value[maxn<<2];
int a[6];
int dp[maxn<<2];
int main() {
    int cas=0;
    while(~scanf("%d%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5])) {
        int sum=0,mid;
        for(int i=0; i<6; i++) {
            sum+=a[i]*(i+1);
        }
        if(!sum)break;
        printf("Collection #%d:\n",++cas);
        int count=0;
        mid=sum>>1;
        if(sum&1) {
            printf("Can't be divided.\n");
            printf("\n");
            continue;
        }
        for(int i=0; i<6; i++) {
            for(int j=1; j<=a[i]; j<<=1) {
                sizes[count]=j*(i+1);
                value[count++]=j*(i+1);
                a[i]-=j;
            }
            if(a[i]>0) {
                sizes[count]=a[i]*(i+1);
                value[count++]=a[i]*(i+1);
            }
        }
        memset(dp,0,sizeof(dp));
        for(int i=0; i<count; i++) {
            for(int j=mid; j>=sizes[i]; j--) {
                dp[j]=max(dp[j],dp[j-sizes[i]]+value[i]);
            }
        }
        if(dp[mid]==mid) {
            printf("Can be divided.\n");
        } else {
            printf("Can't be divided.\n");
        }
        printf("\n");
    }
    return 0;
}

Dividing-多重背包模板题

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原文地址：http://blog.csdn.net/qq_18661257/article/details/47107297

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