Bribing FIPA POJ - 3345 背包类树形DP

M_theory004

于 2019-09-18 14:07:46 发布

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分类专栏：树形DP

本文链接：https://blog.youkuaiyun.com/qq_43313637/article/details/100981533

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本文深入探讨了一种基于深度优先搜索和动态规划的算法，用于解决复杂图结构中的最小成本流问题。该算法通过遍历图中的所有路径，并利用动态规划优化路径选择，以达到最小化总成本的目标。文章详细介绍了算法的设计思路、实现细节及优化策略，为解决实际场景中的物流配送、资源分配等问题提供了有效的解决方案。

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#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<stack>
#include<cstring>
#include<set>
#include<iterator>
#include<list>
#include<deque>
#include<queue>
#include<map>
#include<cmath>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<iomanip>
//#include<unordered_set>
using namespace std;
#define lowbit(x) x&(-x)
typedef long long ll;
typedef pair<int, int> P;
const int N = 2e2 + 10;
const int M = 1e4 + 10;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
vector<int>  g[N];
map<string, int> ma;
int price[N];
int num[N];
int dp[N][N];
int fa[N];
int n, m;
void dfs(int x)
{
	num[x] = 1;
	for (int i = 0; i < g[x].size(); i++)
	{
		int y = g[x][i];
		//cout << x << " " << y << endl;
		dfs(y);
		num[x] += num[y];
	}
	dp[x][0] = 0;
	for (int i = 1; i <= n; i++)
	{
		dp[x][i] = inf;
	}
	dp[x][num[x]] = price[x];
	for (int i = 0; i < g[x].size(); i++)
	{
		int y = g[x][i];
		for (int j = num[x]; j >= 0; j--)
		{
			for (int k = 0; k <= j && k <= num[y]; k++)
			{
				dp[x][j] = min(dp[x][j], dp[x][j - k] + dp[y][k]);
				//printf("dp[%d][%d] == %d\n",x,j, dp[x][j]);
			}
		}
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	string s;
	while (getline(cin,s))
	{
		memset(fa, 0, sizeof(fa));
		if (s == "#")
			break;
		stringstream temps(s);
		temps >> n >> m;
		for (int i = 0; i <= n; i++)
			g[i].clear();
		ma.clear();
		int cnt = 0;
		for (int i = 1; i <= n; i++)
		{
			string name;
			int x;
			getline(cin,s);
			stringstream ss(s);
			ss >> name >> x;
			if (!ma[name])
			{
				ma[name] = ++cnt;
			}
			int u = ma[name];
			price[u] = x;
			while (ss)
			{
				string temp;
				ss >> temp;
				if (temp == "")
					continue;
				if (!ma[temp])
				{
					ma[temp] = ++cnt;
				}
				int v = ma[temp];
				g[u].push_back(v);
				fa[v] = 1;
			}
		}
		price[0] = inf; 
		for (int i = 1; i <= n; i++)
		{
			if (!fa[i])
			{
				g[0].push_back(i);
			}
		}
		dfs(0);
		int ans = inf;
		for (int i = m; i <= n; i++)
		{
			ans = min(ans, dp[0][i]);
			//printf("dp[0][%d] == %d\n", i, dp[0][i]);
		}
		printf("%d\n", ans);
	}
	return 0;
}