UVA 562 Dividing coins （01背包基础）

来源：互联网发布：手机淘宝怎么取消授权编辑：程序博客网时间：2024/06/01 11:53

【题目链接】：click here~~

代码：

/** Problem: UVA No.562* Running time: 0MS* Complier: C++* Author: ACM_herongwei* Create Time: 11:12 2015/9/9 星期三* zeroonebags  * 将金币总价值的一半作为背包容量，然后zeronebags*/#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>#define CLR(c,v) (memset(c,v,sizeof(c)))using namespace std;template <typename _T>inline _T Max(_T a,_T b){    return (a>b)?(a):(b);}template <typename _T>inline _T Maxx(_T a,_T b,_T c){    return (a>Max(b,c))?(a):(Max(b,c));}const int N  =  1e5 + 10;int dp[N];int value[N];int main(){    int Ncase;    scanf("%d",&Ncase);    while(Ncase--)    {        CLR(dp,0);        int sum_cost=0, n_bags;        scanf("%d",&n_bags);        for(int i=0; i<n_bags; ++i)  // max:1000        {            scanf("%d",&value[i]);            sum_cost+=value[i];        }        int mid_cost=sum_cost/2;        for(int i=0; i<n_bags; ++i)        {            for(int j=mid_cost; j>=value[i]; --j)            {                if(dp[j]<=dp[j-value[i]]+value[i])                {                    dp[j]=dp[j-value[i]]+value[i];                }            }        }        printf("%d\n",sum_cost-2*dp[mid_cost]);    }    return 0;}/*sample input332 3 541 2 4 641 4 5 6sample ouput012*/

0 0