HDU 1059.Dividing【多重背包+筛选(DP)】【3月17】
来源:互联网 发布:国泰君安网络金融部 编辑:程序博客网 时间:2024/06/07 09:41
Dividing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22051 Accepted Submission(s): 6208
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 01 0 0 0 1 10 0 0 0 0 0
Sample Output
Collection #1:Can't be divided.Collection #2:Can be divided.
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){ int kase = 1, sum, MAXN; int num[7], dp[60010]; while(1) { sum = 0; for(int i = 1;i <= 6; ++i) { scanf("%d", &num[i]); sum += num[i]*i; } if(sum == 0) break; else if(sum%2 == 1)//总价值为奇数,肯定无法平分 { printf("Collection #%d:\nCan't be divided.\n\n", kase++); continue; } else { printf("Collection #%d:\n", kase++); memset(dp, 0, sizeof(dp)); MAXN = sum/2; //多重背包,这次我们不保存价值,而是看当前价值能否凑出来 for(int i = 0;i <= MAXN && i <= num[1]; ++i) dp[i] = 1; for(int i = 2;i <= 6; ++i) { for(int j = MAXN; j >= 0; --j) { if(dp[j] == 0) continue; for(int k = 1; k <= num[i]; ++k) { if(k*i+j > MAXN) break;//超范围 if(dp[k*i+j] == 1) break;//去重 dp[k*i+j] = 1; } } } if(dp[MAXN]) cout <<"Can be divided.\n\n"; else cout <<"Can't be divided.\n\n"; } } return 0;}
0 0
- HDU 1059.Dividing【多重背包+筛选(DP)】【3月17】
- hdu 1059 Dividing--DP-多重背包问题
- HDU 1059 Dividing(dp多重背包)
- hdu 1059 Dividing (多重背包)
- hdu 1059 Dividing (多重背包 )
- HDU 1059 Dividing(多重背包)
- HDU 1059 Dividing (多重背包)
- HDU Dividing (多重背包+二进制优化)
- HDU 1059 - Dividing(多重背包)
- hdu 1059 Dividing(多重背包)
- HDU 1059 Dividing(多重背包)
- HDU 1059 Dividing(多重背包)
- hdu 1059 Dividing(多重背包)
- HDU-1059 Dividing (多重背包)
- HDU-1059-Dividing(多重背包)
- POJ1014 Dividing 多重背包[DP]
- hdu 1059 Dividing DP,多重背包 测试数据很水
- HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化)
- 《第一行代码》读书笔记
- Missing contentDescription attribute on image
- C语言习题 矩阵元素变换
- x264代码剖析(十三):核心算法之帧间预测函数x264_mb_analyse_inter_*()
- 2597: [Wc2007]剪刀石头布 费用流
- HDU 1059.Dividing【多重背包+筛选(DP)】【3月17】
- Paper Reading 2:Human-level control through deep reinforcement learning
- ubuntu14.04+opencv 3.0安装及测试
- UESTC 483 Data Structure Problem
- golang入门-- 一个2D的图形库学习
- 史上最详细的Android Studio系列教程四--Gradle基础
- java多线程之中断(interrupt)问题
- 第八章:JavaScript事件驱动编程和访问CSS技术
- Oracle SQL语句执行顺序