808E. Selling Souvenirs***

最新推荐文章于 2020-07-27 09:16:49 发布

AndrewThompson

最新推荐文章于 2020-07-27 09:16:49 发布

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分类专栏： Codeforces

本文链接：https://blog.youkuaiyun.com/zju2016/article/details/72587260

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本文介绍了一种解决复杂背包问题的算法实现，通过定义结构体dp来存储价值和数量信息，并利用动态规划的方法来求得最大价值。文章通过具体代码展示了如何对不同类型的物品进行排序和累积价值计算，最终输出最优解。

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#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<algorithm>
#include<queue>
#include<list>
#include<stack>
#include<cstdio>
#include<fstream>
#include<numeric>
#include<functional>
#include<utility>
#include<memory>
#include<iomanip>
using namespace std;

typedef long long int LL;

LL n, m;

typedef struct dp{
	LL val;
	LL number1, number2;
}dp;


int main(){

	while (cin >> n >> m){
		LL value[4][300010];
		LL total[4][300010];
		vector<LL> index(4, 0);
		for (LL i = 0; i < n; i++){
			LL w, c;
			cin >> w >> c;
			index[w]++;
			value[w][index[w]] = c;
		}
		for (LL i = 1; i <= 3; i++){
			sort(value[i] + 1, value[i] + index[i] + 1, greater<LL>());
			total[i][0] = 0;
			for (LL j = 1; j <= index[i]; j++) total[i][j] = total[i][j - 1] + value[i][j];
		}
		dp result[300010];
		result[0].val = 0;
		result[0].number1 = 0;
		result[0].number2 = 0;
		for (LL i = 1; i <= m; i++){
			result[i] = result[i - 1];
			if (result[i - 1].val + value[1][result[i - 1].number1 + 1]>result[i].val){
				result[i].val = result[i - 1].val + value[1][result[i - 1].number1 + 1];
				result[i].number2 = result[i - 1].number2;
				result[i].number1 = result[i - 1].number1 + 1;
			}
			if (i >= 2 && result[i - 2].val + value[2][result[i - 2].number2 + 1] > result[i].val){
				result[i].val = result[i - 2].val + value[2][result[i - 2].number2 + 1];
				result[i].number1 = result[i - 2].number1;
				result[i].number2 = result[i - 2].number2 + 1;
			}
		}
		LL res = 0;
		for (LL i = 0; i <= index[3]; i++){
			if (m >= i * 3){
				res = max(res, total[3][i] + result[m - 3 * i].val);
			}
		}
		cout << res << endl;
	}
	return 0;
}