01背包记录路径)
来源:互联网 发布:写java用什么eclipse 编辑:程序博客网 时间:2024/05/24 06:58
题目信息
1068. Find More Coins (30)
时间限制150 ms
内存限制65536 kB
代码长度限制16000 B
Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 104 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10^4, the total number of coins) and M(<=10^2, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the face values V1 <= V2 <= … <= Vk such that V1 + V2 + … + Vk = M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output “No Solution” instead.
Note: sequence {A[1], A[2], …} is said to be “smaller” than sequence {B[1], B[2], …} if there exists k >= 1 such that A[i]=B[i] for all i < k, and A[k] < B[k].
Sample Input 1:
8 9
5 9 8 7 2 3 4 1
Sample Output 1:
1 3 5
Sample Input 2:
4 8
7 2 4 3
Sample Output 2:
No Solution
#include<iostream>#include<algorithm>#include<cstring>#include<vector>#define LL long long#define INF 0x3f3f3f3fusing namespace std;const int maxn=10005;const int maxv=105;int coins[maxn];int dp[maxv];bool flag[maxn][maxv];//表示背包容量为j时,是否含有第i个物品 int ans[maxn];int cnt=0;bool cmp(int a,int b){return a>b;}int main(){int n,m;scanf("%d %d",&n,&m);for(int i=0;i<n;i++)scanf("%d",&coins[i]);sort(coins,coins+n,cmp);//按从大到小排序,即背包里先放大的,再放小的for(int i=0;i<n;i++){for(int j=m;j>=coins[i];j--){if(dp[j]<=dp[j-coins[i]]+coins[i]){dp[j]=dp[j-coins[i]]+coins[i];flag[i][j]=true;} }}if(dp[m]!=m)printf("No Solution");else{int v=m,idx=n-1;while(v){if(flag[idx][v]){ans[cnt++]=coins[idx];v-=coins[idx];}idx--;}printf("%d",ans[0]);for(int i=1;i<cnt;i++)printf(" %d",ans[i]);}return 0;}
- 01背包记录路径)
- 01背包+记录路径
- Uva624(记录路径的01背包)
- uva 01背包记录路径
- NOJ1308 背包问题 (背包记录路径)
- 新年趣事之打牌(01背包+记录路径)
- UVA 624 CD(01背包+记录路径)
- UVA 624 CD(01背包/记录路径)
- CD - UVa 624 01背包记录路径
- 01背包状态压缩和记录路径
- 记录路径的01背包问题
- POJ_1787_Charlie'sChange(多重背包&&记录路径)
- 记录路径的背包问题
- 背包问题加记录路径
- 0/1背包记录路径
- 01背包、完全背包、多重背包问题的C++实现及路径记录
- 2017百度之星资格赛:1004. 度度熊的午饭时光(01背包+记录路径)
- hdu2126Buy the souvenirs (01背包+记录路径的种数)
- [BZOJ3930][CQOI2015]选数(数论+容斥)
- linux下的find文件查找命令与grep文件内容查找命令
- IDEA 重复代码快速重构
- EventBus总结
- Java 并发编程 之 volatile(三)
- 01背包记录路径)
- 宽度已知,实现三栏布局的方案(5种layout方案)
- Linux服务器Zookeeper+Dubbo环境搭建
- java基础语法之自加(++)自减(--)讲解-----看了就会!!!
- vue.js开发外卖App项目总结(二)
- rapidjson将map转为json------人生苦短,我用rapidjson
- 位压缩
- oracle--认识表与约束
- 单元测试多线程输出问题