hdu 1171 Big Event in HDU

    xiaoxiao2021-12-12  8

    Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002. 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).   Input Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the 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.   Output For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.   Sample Input 2 10 1 20 1 3 10 1 20 2 30 1 -1   Sample Output 20 10

    40 40

    题意:就是说要分家了,这么些个设备,每个设备有一定的价值(把相同价值的设备当成同一种),汝瑰哦能平分的话,就平分,不能平分就A的多分。

    思路:AB分的话,就A>=B,那么sum/2就是小于等于总数的一半,对于B,把sum/2当作背包,装每种物品,得到最大的价值。

    代码:

    #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int main() {   int value[5005];   int dp[125005];   int n,m,v,sum,i,j,l;   while(scanf("%d",&n)&&n>0)   {    memset(dp,0,sizeof(dp));    memset(value,0,sizeof(value));    l=0;    sum=0;    while(n--)     {       scanf("%d%d",&v,&m);       sum+=(v*m);       while(m--)//因为01背包是每件物品只有一件,考虑装不装       value[l++]=v;           }     for(i=0;i<l;i++)//把每件物品都试一遍     {         for(j=sum/2;j>=value[i];j--)//针对i物品,每个价值的空闲都试一遍             dp[j]=max(dp[j],dp[j-value[i]]+value[i]);     }     printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);   } }

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