第五届程序设计大赛 Divideing Jewels(多重背包)
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Divideing Jewels
Mary and Rose own a collection of jewells. They want to split the collection among themselves so that both receive an equal share of the jewels. This would be easy if all the jewels had the same value, because then they could just split the collection in half. But unfortunately, some of the jewels are larger, or more beautiful than others. So, Mary and Rose start by assigning a value, a natural number between one and ten, to each jewel. Now they want to divide the jewels so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the jewels in this way (even if the total value of all jewels is even). For example, if there are one jewel 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 jewels.
【Standard input】
Each line in the input file describes one collection of jewels to be divided. The lines contain ten non-negative integers n1 , . . . , n10 , where ni is the number of jewels of value i. The maximum total number of jewells will be 10000.
The last line of the input file will be "0 0 0 0 0 0 0 0 0 0"; do not process this line.
【Standard output】
For each collection, output "#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.
【Standard input】
1 0 1 2 0 0 0 0 2 0
1 0 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
【Standard output】
#1:Can't be divided.
#2:Can be divided.
多重背包 竟然超时
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
int a[11],dp[5010];
int main()
{
int x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,g[5010],count=1;
while(scanf("%d%d%d%d%d%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5,&x6,&x7,&x8,&x9,&x10),x1||x2||x3||x4||x5||x6||x7||x8||x9||x10){
a[1]=x1;a[2]=x2;a[3]=x3;a[4]=x4;a[5]=x5;a[6]=x6;a[7]=x7;a[8]=x8;a[9]=x9;a[10]=x10;
int cn=0,sum=0;
for(int i=1;i<=10;i++){
int m=a[i];
if(m){
sum+=m*i;
for(int j=1;j<=m;j<<=1){
g[cn++]=j*i;
m-=j;
}
if(m>0)g[cn++]=m*i;
}
}
int div=sum/2;
memset(dp,0,sizeof(dp));
for(int i=0;i<cn;i++)
for(int j=div;j>=0;j--){
if(g[i]<=j)dp[j]=max(dp[j],dp[j-g[i]]+g[i]);
}
int A=sum-dp[div];
if(A==dp[div])printf("#%d:Can be divided.\n\n",count++);
else printf("#%d:Can't be divided.\n\n",count++);
}
}
DFS
#include <stdio.h>
#include <string.h>
int sum,a[11],flag;
void dfs(int n)
{
int i;
if(flag)
return;
if(n==sum)
{
flag=1;
return;
}
for(i=10;i>=1;i--)
{
if(a[i]>0&&n+i<=sum)
{
a[i]--;
dfs(n+i);
if(flag)
return;
}
}
}
int main()
{
int i,t;
for(t=1;;t++)
{
sum=0;
for(i=1;i<=10;i++)
{
scanf("%d",&a[i]);
sum+=a[i]*i;
}
if(sum==0)
break;
if(t!=1)
printf("\n");
if(sum%2==1)
{
printf("#%d:Can't be divided.\n",t);
continue;
}
sum/=2;
flag=0;
dfs(0);
if(flag)
printf("#%d:Can be divided.\n",t);
else
printf("#%d:Can't be divided.\n",t);
}
}
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