【二进制拆分多重背包】【HDU1059】【Dividing】
Dividing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18190 Accepted Submission(s): 5080
Problem 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.
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.
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.
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.
Source
Mid-Central European Regional Contest 1999
一开始还觉得有点难写。。
后来AC才感觉自己编码能力已经很强了。。
主要是这个拆分函数就OK了
void solve()
{
for(int i=1;i<=6;i++)
{
int temp=0;
while(n[i]>0)
{
if(n[i]>=ER[temp])
A[++tot]=i*ER[temp];
else
A[++tot]=i*n[i];
n[i]=n[i]-A[tot]/i;
temp++;
}
}
}
完整代码如下：
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
#define MAX 2100000000
using namespace std;
void init()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
int n[10];
int F[100001];
int ER[100];
int K;
int A[600];
int tot=0; //总数
void YCL()
{
memset(F,0,sizeof(F));
tot=0;
K=0;
for(int i=1;i<=6;i++)
{
K+=i*n[i];
}
for(int i=1;i<=K/2;i++)
F[i]=MAX;
F[0]=0;
}
void ER1()
{
ER[0]=1;
for(int i=1;i<=15;i++)
{
ER[i]=ER[i-1]*2;
}
}
void solve()
{
for(int i=1;i<=6;i++)
{
int temp=0;
while(n[i]>0)
{
if(n[i]>=ER[temp])
A[++tot]=i*ER[temp];
else
A[++tot]=i*n[i];
n[i]=n[i]-A[tot]/i;
temp++;
}
}
}
int main()
{
// init();
int CASE=0;
ER1(); //得到二的倍数
while(cin>>n[1]>>n[2]>>n[3]>>n[4]>>n[5]>>n[6]&&(n[1]||n[2]||n[3]||n[4]||n[5]||n[6]))
{
CASE++;
YCL(); //计算K
solve(); //拆分开始
if(K%2==0)
{
for(int i=1;i<=tot;i++)
for(int j=K/2;j>=0;j--)
{
if(j-A[i]>=0)
F[j]=min(F[j],F[j-A[i]]);
}
printf("Collection #%d:\n",CASE);
if(F[K/2]!=MAX)
{
printf("Can be divided.\n\n");
}
else
{
printf("Can't be divided.\n\n");
}
}
else
{
printf("Collection #%d:\n",CASE);
printf("Can't be divided.\n\n");
}
}
return 0;
}
恩 再补一道同类型的题目
HDU（3732）
Ahui Writes Word
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2214 Accepted Submission(s): 811
Problem Description
We all know that English is very important, so Ahui strive for this in order to learn more English words. To know that word has its value and complexity of writing (the length of each word does not exceed 10 by only lowercase letters), Ahui wrote the complexity of the total is less than or equal to C.
Question: the maximum value Ahui can get.
Note: input words will not be the same.
Question: the maximum value Ahui can get.
Note: input words will not be the same.
Input
The first line of each test case are two integer N , C, representing the number of Ahui’s words and the total complexity of written words. (1 ≤ N ≤ 100000, 1 ≤ C ≤ 10000)
Each of the next N line are a string and two integer, representing the word, the value(Vi ) and the complexity(Ci ). (0 ≤ Vi , Ci ≤ 10)
Each of the next N line are a string and two integer, representing the word, the value(Vi ) and the complexity(Ci ). (0 ≤ Vi , Ci ≤ 10)
Output
Output the maximum value in a single line for each test case.
Sample Input
5 20
go 5 8
think 3 7
big 7 4
read 2 6
write 3 5
Sample Output
15
Hint
Input data is huge,please use “scanf(“%s”,s)”
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
int N,C;
char buffer[30];
int map[20][22];
int ER[30];
int tot;
int v[3000];
int w[3000];
int F[10001];
void ER1()
{
ER[0]=1;
for(int i=1;i<=20;i++)
{
ER[i]=ER[i-1]*2;
}
}
void input()
{
int a,b;
tot=0;
memset(map,0,sizeof(map));
memset(F,0,sizeof(F));
for(int i=1;i<=N;i++)
{
scanf("%s %d %d",buffer,&a,&b);
map[a][b]++;
}
}
void init()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
void solve()
{
for(int i=1;i<=10;i++)
for(int j=0;j<=10;j++)
{
int temp=0;
while(map[i][j]>0)
{
if(map[i][j]>=ER[temp])
{
v[++tot]=ER[temp]*i;
w[tot]=ER[temp]*j;
}
else
{
v[++tot]=map[i][j]*i;
w[tot]=map[i][j]*j;
}
{
map[i][j]=map[i][j]-v[tot]/i;
temp++;
}
}
}
}
void dp()
{
for(int i=1;i<=tot;i++)
for(int j=C;j>=0;j--)
{
if(j-w[i]>=0)
F[j]=max(F[j],F[j-w[i]]+v[i]);
}
}
int main()
{
// init();
ER1();
while(cin>>N>>C)
{
input();
solve();
dp();
printf("%d\n",F[C]);
}
return 0;
}