[贪心&&排序]UVA10670 Work Reduction

Problem C: Work Reduction

The only hope for you now is to hire help. There are various agencies which offer paperwork reduction plans:

For $A they will reduce your paperwork by one unit.
For $B they will reduce your entire paperwork by half (rounding down when necessary).

Note that work can never be reduced to less than 0.

Your task now is to produce a sorted table of agency names and their respective minimum costs to solve your workload problem.

The first line of input consists of a single positive integer representing the number of cases to follow. Each case begins with three positive integers separated by spaces: N - your starting workload, M - your target workload, and L - the number of work reduction agencies available to you, (1 <= M <= N <= 100000, 1 <= L <= 100). The next L lines have the format "[agency name]:A,B", whereA and B are the rates as described above for the given agency. (0 <= A,B <= 10000) The length of the agency name will be between 1 and 16, and will consist only of capital letters. Agency names will be unique.

For each test case, print "Case X", with X being the case number, on a single line, followed by the table of agency names and their respective minimum costs, sorted in non-decreasing order of minimum costs. Sort job agencies with identical minimum costs in alphabetical order by agency name. For each line of the table, print out the agency name, followed by a space, followed by the minimum required cost for that agency to solve your problem.

Sample Input

2
100 5 3
A:1,10
B:2,5
C:3,1
1123 1122 5
B:50,300
A:1,1000
C:10,10
D:1,50
E:0,0

Sample Output

Case 1
C 7
B 22
A 37
Case 2
E 0
A 1
D 1
C 10
B 50

题意：现在有N个任务要做，而且要留M个到明天做，你自己做不完，只能请团队帮你做，每个团队有两种方案，一个一个任务做或者把你当前的任务的一半做掉，花费各不相同，求出对每一个团队的最小花费。

思路：典型的贪心题目，每次都选择最优方案，并且保证当前的任务数量大于等于M。

代码：

#include<iostream>
#include<algorithm>
#include<string>

using namespace std;

class Node
{
public:
	string str;
	int cost;
}node[10005];

bool cmp(Node s1,Node s2)
{
	if(s1.cost!=s2.cost) return s1.cost<s2.cost;
	else return s1.str<s2.str;
}

int main()
{
	int num,pos;
	cin>>num;
	for(pos=1;pos<=num;pos++)
		{
			int x,y,z,r,t;
			cin>>x>>y>>z;
			char ch;
			for(int i=0;i<z;i++)
				{
					int work=x;
					string st="";
					while(cin>>ch&&ch!=‘:‘)
						{
							st=st+ch;
						}
					node[i].str=st;
					cin>>r>>ch>>t;
					int val=0;
					while(t<((work+1)/2*r)&&(work-(work+1)/2)>=y)
						{
							val=val+t;
							work=work-(work+1)/2;
						}
					val=val+(work-y)*r;
					node[i].cost=val;
				}
			sort(node,node+z,cmp);
			cout<<"Case "<<pos<<endl;
			for(int i=0;i<z;i++)
				{
					cout<<node[i].str<<" "<<node[i].cost<<endl;
				}
		}
	return 0;
}

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[贪心&&排序]UVA10670 Work Reduction

原文：http://blog.csdn.net/zju_ziqin/article/details/23022869