A1018 Public Bike Management （30 分）

最新推荐文章于 2021-03-12 10:47:37 发布

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最新推荐文章于 2021-03-12 10:47:37 发布

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分类专栏：算法杂题

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本文介绍了一种结合最短路径算法与资源优化策略的算法实现，通过Dijkstra算法寻找两点间的最短路径，并在此基础上优化资源分配，减少携带和搬运成本。该算法适用于需要在路径选择和资源消耗间做权衡的场景。

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 510;
const int INF = 1e9;
int cmax, n, sp, m;
int num[maxn] = { 0 }, d[maxn], vis[maxn] = { 0 };
int G[maxn][maxn];
vector<int> pre[maxn],temp,path;
void dijkstra(int s)
{
	fill(d, d + maxn, INF);
	d[s] = 0;
	for (int i = 0; i <= n; i++)
	{
		int u = -1, min = INF;
		for (int j = 0; j <= n; j++)
		{
			if (vis[j] == 0 && d[j] < min)
			{
				u = j;
				min = d[j];
			}
		}
		if (u == -1) return;
		vis[u] = 1;
		for (int v = 0; v <= n; v++)
		{
			if (vis[v] == 0 && G[u][v] != INF)
			{
				if (d[u] + G[u][v] < d[v])
				{
					pre[v].clear();
					d[v] = d[u] + G[u][v];
					pre[v].push_back(u);
				}
				else if (d[u] + G[u][v] == d[v]) pre[v].push_back(u);
			}
		}
	}
}
int minbring = INF, mintake = INF;
void dfs(int v)
{
	int bring = 0, take = 0;
	if (v == 0)
	{
		temp.push_back(v);
		for (int i = temp.size() - 2; i >= 0; i--)
		{
			int id = temp[i];
			if (num[id] > 0) take += num[id];
			else
			{
				if (abs(num[id]) > take)
				{
					bring += abs(num[id]) - take;
					take = 0;
				}
				else
				{
					take -= abs(num[id]);
				}
			}
		}
		if (bring < minbring)
		{
			minbring = bring;
			mintake = take;
			path = temp;
		}
		else if (bring == minbring && take < mintake)
		{
			mintake = take;
			path = temp;
		}
		temp.pop_back();
		return;
	}
	temp.push_back(v);
	for (int i = 0; i < pre[v].size(); i++)
	{
		dfs(pre[v][i]);
	}
	temp.pop_back();
}
int main()
{
	int a, b, t;
	fill(G[0], G[0] + maxn*maxn, INF);
	scanf("%d%d%d%d", &cmax, &n, &sp, &m);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &num[i]);
		num[i] = num[i] - 0.5*cmax;
	}
	for (int i = 0; i < m; i++)
	{
		scanf("%d%d%d", &a, &b, &t);
		G[a][b] = G[b][a] = t;
	}
	dijkstra(0);
	dfs(sp);
	printf("%d ", minbring);
	for (int i = path.size() - 1; i >= 0; i--)
	{
		printf("%d", path[i]);
		if (i != 0) printf("->");
	}
	printf(" %d", mintake);
	return 0;
}