HDU1171--01背包
Big Event in HDU
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36096 Accepted Submission(s): 12533
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is
N (0<N<1000) kinds of facilities (different value, different kinds).
facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.
2
10 1
20 1
3
10 1
20 2
30 1
-1
20 10
40 40
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<deque>
#include<map>
#include<set>
#include<algorithm>
#include<string>
#include<iomanip>
#include<cstdlib>
#include<cmath>
#include<sstream>
#include<ctime>
using namespace std; int va[1275005];//(50+1)*50/2*100=1275000最大价值作为背包的最大空间
int dp[1275005]; int main()
{
int n;
int i,j;
int v,num;//价值,数量
int count_num;//所有物品的数量
int count_pac;//总的背包大小
while(scanf("%d",&n) != EOF)
{
if(n < 0)
{
return 0;
}
memset(va,0,sizeof(va));
memset(dp,0,sizeof(dp));
j = 1;
count_num = 0;
count_pac = 0;
for(i = 1; i <= n; i++)
{
scanf("%d%d",&v,&num);
count_num += num;
count_pac += num * v;
while(num--)
{
va[j] = v;
j++;
}
}
//01背包
for(i = 1; i <= count_num; i++)
{
for(j = count_pac / 2; j >= va[i]; j--)
{
dp[j] = max(dp[j], dp[j - va[i]] + va[i]);
}
}
printf("%d %d\n",count_pac - dp[count_pac / 2], dp[count_pac / 2]);
}
return 0;
}
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