hihoCoder 1137-Recruitment

hihoCoder 1137-Recruitment

#1137 : Recruitment
时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述
A company plans to recruit some new employees. There are N candidates (indexed from 1 to N) have taken the recruitment examination. After the examination, the well-estimated ability value as well as the expected salary per year of each candidate is collected by the Human Resource Department.

Now the company need to choose their new employees according to these data. To maximize the company's benefits, some principles should be followed:

1. There should be exactly X males and Y females.

2. The sum of salaries per year of the chosen candidates should not exceed the given budget B.

3. The sum of ability values of the chosen candidates should be maximum, without breaking the previous principles. Based on this, the sum of the salary per year should be minimum.

4. If there are multiple answers, choose the lexicographically smallest one. In other words, you should minimize the smallest index of the chosen candidates; If there are still multiple answers, then minimize the second smallest index; If still multiple answers, then minimize the third smallest one; ...

Your task is to help the company choose the new employees from those candidates.

输入
The first line contains four integers N, X, Y, and B, separated by a single space. The meanings of all these variables are showed in the description above. 1 <= N <= 100, 0 <= X <= N, 0 <= Y <= N, 1 <= X + Y <= N, 1 <= B <= 1000.

Then follows N lines. The i-th line contains the data of the i-th candidate: a character G, and two integers V and S, separated by a single space. G indicates the gender (either "M" for male, or "F" for female), V is the well-estimated ability value and S is the expected salary per year of this candidate. 1 <= V <= 10000, 0 <= S <= 10.

We assure that there is always at least one possible answer.

输出
On the first line, output the sum of ability values and the sum of salaries per year of the chosen candidates, separated by a single space.

On the second line, output the indexes of the chosen candidates in ascending order, separated by a single space.

样例输入
4 1 1 10
F 2 3
M 7 6
M 3 2
F 9 9

样例输出
9 9
1 2


本题实质上是一个0/1背包问题,将男性和女性分别背包。

以男性为例,转换为从若干个(n1)人中取出X个男性,保证他们的工资总和不超过B,但最大化能力总和。每个应聘者相当于一个物品,工资总和为B相当于容量为B的背包。常规的0/1背包是从这n1个应聘者中挑任意多个,使得工资总和不超过B,但能力总和最大。但是本题的背包限定了取且只取X个应聘者,使得工资总和不超过B,但能力总和最大。

DP的状态转移过程为:假设dpm[i][j]表示从【当前】所有男性中选取i个,工资总和为j时获得的最大能力总和。

假设X=5,当前正好来了5个男性,已经求到了dpm[X][j],当再来一位男性应聘者a(能力为v,工资为s)时,dpm[X][j]含义就是从当前6位应聘者中选5位满足限定条件。此时有两个选择,要或者不要a,如果不要a,则dpm[X][j]结果不变;如果要a,则前5个只能取4个,加上a就是5个,此时能力总和就是dpm[X-1][j-s]+v,所以最终dpm[X][j]=max{dpm[X][j], dpm[X-1][j-s]+v}。

当把所有男性应聘者都测试过之后,dpm[X][j]就是从所有n1个男性中选X个,其工资总和为j时的能力总和。

最后把男性和女性的情况一组合,再满足总预算不超过B,取全局最优结果。

因为应聘者N最多为100,所以使用数组unsigned long long id[2]来记录应聘者选取情况,unsigned long long 为64位,每位二进制代表一名应聘者,为1选中,id[0]记录前50位,id[1]记录后50位。

完整代码如下:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int kMaxN = 105, kMaxB = 1005;
int dpf[kMaxN][kMaxB], dpm[kMaxN][kMaxB];//dpf[i][j]从所有女性应聘者中选取i个,工资总和为j时的最大能力和
unsigned long long idf[kMaxN][kMaxB][2];//记录女性选取情况
unsigned long long idm[kMaxN][kMaxB][2];//记录男性选取情况
int N, X, Y, B, V, S;
char G[2];
int ans_v = 0, ans_s = 0;//总能力,总工资
unsigned long long ans_id[2] = { 0 };//总选取情况
void DP()
{
	dpf[0][0] = dpm[0][0] = 0;
	int cnt_f = 0, cnt_m = 0, sum_s = 0;
	for (int i = 1; i <= N; i++)
	{
		scanf("%s %d %d", G, &V, &S);
		sum_s += S;
		sum_s = min(sum_s, B);
		if (G[0] == 'F')
		{
			cnt_f++;
			cnt_f = min(cnt_f, Y);//最多选取Y位
			for (int j = cnt_f; j >= 1; j--)
			{
				for (int k = sum_s; k >= S; k--)
				{
					if (dpf[j - 1][k - S] < 0)
						continue;
					if (dpf[j - 1][k - S] + V > dpf[j][k])
					{
						dpf[j][k] = dpf[j - 1][k - S] + V;
						idf[j][k][0] = idf[j - 1][k - S][0];
						idf[j][k][1] = idf[j - 1][k - S][1];
						if (i <= 50)
							idf[j][k][0] |= 1LL << (i - 1);//选中第i位
						else
							idf[j][k][1] |= 1LL << (i - 1 - 50);//选中第i位
					}
				}
			}
		}
		else
		{
			cnt_m++;
			cnt_m = min(cnt_m, X);
			for (int j = cnt_m; j >= 1; j--)
			{
				for (int k = sum_s; k >= S; k--)
				{
					if (dpm[j - 1][k - S] < 0)
						continue;
					if (dpm[j - 1][k - S] + V > dpm[j][k])
					{
						dpm[j][k] = dpm[j - 1][k - S] + V;
						idm[j][k][0] = idm[j - 1][k - S][0];
						idm[j][k][1] = idm[j - 1][k - S][1];
						if (i <= 50)
							idm[j][k][0] |= 1LL << (i - 1);
						else
							idm[j][k][1] |= 1LL << (i - 1 - 50);
					}
				}
			}
		}
	}
}
void Match()
{
	for (int i = 0; i <= B; i++)
	{
		if (dpf[Y][i] == -1)
			continue;
		for (int j = 0; j + i <= B; j++)
		{
			if (dpm[X][j] == -1)
				continue;
			if (dpf[Y][i] + dpm[X][j] > ans_v)//能力更强
			{
				ans_v = dpf[Y][i] + dpm[X][j];
				ans_s = i + j;
				ans_id[0] = idf[Y][i][0] | idm[X][j][0];
				ans_id[1] = idf[Y][i][1] | idm[X][j][1];
			}
			else if (dpf[Y][i] + dpm[X][j] == ans_v && (i + j) < ans_s)//能力相同,但工资更少
			{
				ans_s = i + j;
				ans_id[0] = idf[Y][i][0] | idm[X][j][0];
				ans_id[1] = idf[Y][i][1] | idm[X][j][1];
			}
			else if (dpf[Y][i] + dpm[X][j] == ans_v && (i + j) == ans_s)//能力和工资都相同
			{
				int id0 = idf[Y][i][0] | idm[X][j][0];
				int id1 = idf[Y][i][1] | idm[X][j][1];
				if (ans_id[0] > id0)//排序靠前
				{
					ans_id[0] = id0;
					ans_id[1] = id1;
				}
				else if (ans_id[1] > id1)//排序靠前
				{
					ans_id[0] = id0;
					ans_id[1] = id1;
				}
			}
		}
	}
}
void FormatOut()
{
	printf("%d %d\n", ans_v, ans_s);
	for (int i = 1; i <= 50; i++)
	{
		if (ans_id[0] & 1)
			printf("%d ", i);
		ans_id[0] >>= 1;
	}
	for (int i = 51; i <= 100; i++)
	{
		if (ans_id[1] & 1)
			printf("%d ", i);
		ans_id[1] >>= 1;
	}
}
int main()
{
	memset(dpf, -1, sizeof(dpf));
	memset(dpm, -1, sizeof(dpm));
	memset(idf, 0, sizeof(idf));
	memset(idm, 0, sizeof(idm));
	scanf("%d %d %d %d", &N, &X, &Y, &B);
	DP();
	Match();
	FormatOut();
 	return 0;
}

代码大部分参考此博客,感谢。

本代码提交AC,用时33MS,内存4MB。

Leave a Reply

Your email address will not be published. Required fields are marked *