H - I love exam (HDU-6968)

这是一个典型的背包问题 让我们求学生Z在t天挂科数小于等于p的情况下借助总共给出的m份资料所得的最大分数总和

我们用 f[i][j] 来表示第i科花费j时间所得的最大分数
用 dp[i][j][k] 来表示前i门科目花费j时间挂科数为k的情况下所得的最大分数总和

求f 和 dp都是01背包 只不过一个是科目得分 一个是 f 作为价值

对于字符串我们可以用map转化为序号id 从而方便我们存数 再设一个vector来保存所有当前序号的价值和花费时间

1. f[i][j] 的修改

for (int i = 1; i <= n; i ++ )
{
	f[i][0] = 0;
	for (int j = 0; j < cour[i].size(); j ++ )
	{
		int val = cour[i][j].val, time = cour[i][j].time;
		for (int k = t; k >= time; k -- )
			f[i][k] = max(f[i][k], f[i][k - time] + val);
	}
}

2. dp[i][j][k] 的修改

for (int i = 1; i <= n; i ++ )
{
	for (int j = 0; j <= t; j ++ ) // 当前科目花费时间
	for (int k = j; k <= t; k ++ ) // 总时间
	for (int lim = 0; lim <= p; lim ++ ) // 挂科数
	{
		if (f[i][j] < 0) continue;
		int flag = 0;
		if (f[i][j] < 60) flag = 1;
		if (f[i][j] > 100) f[i][j] = 100;
		if (lim >= flag)
			dp[i][k][lim] = max(dp[i][k][lim], dp[i - 1][k - j][lim - flag] + f[i][j]);
	}
}

3. 总代码

#include 

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII; // x-val y-time

const int N = 60, M = 3010;

int n, m, t, p;
int f[N][M], dp[N][M][10];
map<string, int> mp;
vector<PII> course[N];

int main()
{
	int T;
	scanf("%d", &T);
	
	while (T -- )
	{
		scanf("%d", &n);		
		mp.clear();
		for (int i = 1; i <= n; i ++ ) course[i].clear();
		for (int i = 1; i <= n; i ++ )
		{
			string s;
			cin >> s;
			mp[s] = i;
		}
		
		scanf("%d", &m);
		for (int i = 0; i < m; i ++ )
		{
			string s;
			cin >> s;
			int id = mp[s];
			PII a;
			cin >> a.x >> a.y;
			course[id].push_back(a);
		}
		
		scanf("%d%d", &t, &p);
		
		memset(f, -0x3f, sizeof f);
		memset(dp, -0x3f, sizeof dp);
		
		dp[0][0][0] = 0;
		
		for (int i = 1; i <= n; i ++ )
		{
			f[i][0] = 0;
			for (int j = 0; j < course[i].size(); j ++ )
			{
				int val = course[i][j].x, time = course[i][j].y;
				for (int k = t; k >= time; k -- )
					f[i][k] = max(f[i][k], f[i][k - time] + val);
			}
			
			for (int j = 0; j <= t; j ++ )
			for (int k = j; k <= t; k ++ )
			for (int lim = 0; lim <= p; lim ++ )
			{
				int flag = 0;
				if (f[i][j] < 0) continue;
				if (f[i][j] < 60) flag = 1;
				if (f[i][j] > 100) f[i][j] = 100;
				if (lim >= flag)
					dp[i][k][lim] = max(dp[i][k][lim], dp[i - 1][k - j][lim - flag] + f[i][j]);
			}
		}
		
		int ans = -1;
		for (int i = 0; i <= t; i ++ )
			for (int j = 0; j <= p; j ++ )
				ans = max(ans, dp[n][i][j]);
		if (ans < 0) puts("-1");
		else printf("%d\n", ans);
	}
	
	return 0;
}