Light OJ - 1002 - Country Roads 题解

变形Dijkstra算法解析

最新推荐文章于 2019-08-12 10:54:55 发布

最新推荐文章于 2019-08-12 10:54:55 发布 · 198 阅读

本文介绍了一种变形的Dijkstra算法，该算法用于寻找图中从各节点到指定节点的路径，路径成本由所经过边的最大权值决定。文章通过一个具体的示例详细解释了算法的实现过程，并给出了完整的C++代码。

不是很难的题目，却搞得我好头痛，这里是应该了变形的Dijstra算法了，因为这里求最短路径不是把所有路径相加，而是把路径上所有的值取最大值，这个最大值越小，就说明结果最优，也不管你绕了多长的路。

另外值得注意的就是memset这个函数，看来很多人栽跟斗啊，这个函数只能用来清零，是不能用来初始化值的，因为它是按照字符来初始化值的，所以计算好也可以给整形清零，但是这次我用来给bool数组清零，结果出现奇怪的现象。还是使用fill这个函数或者是循环赋值吧，不能偷懒的了。

下面是题目：

I am going to my home. There are many cities and many bi-directional roads between them. The cities are numbered from0ton-1and each road has a cost. There aremroads. You are given the number of my citytwhere I belong. Now from each city you have to find the minimum cost to go to my city. The cost is defined by the cost of the maximum road you have used to go to my city.

For example, in the above picture, if we want to go from 0 to 4, then we can choose

1)0 - 1 - 4 which costs 8, as 8 (1 - 4) is the maximum road we used

2)0 - 2 - 4 which costs 9, as 9 (0 - 2) is the maximum road we used

3)0 - 3 - 4 which costs 7, as 7 (3 - 4) is the maximum road we used

So, our result is 7, as we can use 0 - 3 - 4.

Input

Input starts with an integerT (≤ 20), denoting the number of test cases.

Each case starts with a blank line and two integersn (1 ≤ n ≤ 500)andm (0 ≤ m ≤ 16000). The nextmlines, each will contain three integersu, v, w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 20000)indicating that there is a road betweenuandvwith costw. Then there will be a single integert (0 ≤ t < n). There can be multiple roads between two cities.

Output

For each case, print the case number first. Then for all the cities (from0ton-1) you have to print the cost. If there is no such path, print'Impossible'.

Sample Input

Output for Sample Input

5 6

0 1 5

0 1 4

2 1 3

3 0 7

3 4 6

3 1 8

5 4

0 1 5

0 1 4

2 1 3

3 4 7

Case 1:

Case 2:

Impossible

题目还是挺有意思的，变形的Dijstra，学好这个算法还是很有必要的。

标准的Dijstra参考资料：

http://www.geeksforgeeks.org/greedy-algorithms-set-6-dijkstras-shortest-path-algorithm/

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <string>
#include <vector>
#include <assert.h>

using namespace std;

const static int MAX_COST = 20001;

int minDist(vector<int> &rs, bool *splitSet)
{
	int min_val = MAX_COST, id = -1;
	for (unsigned i = 0; i < rs.size(); i++)
	{
		if (!splitSet[i] && rs[i] < min_val)
		{
			min_val = rs[i];
			id = i;
		}
	}
	return id;
}

vector<int> findPaths(vector<vector<int> > &graph, const int sp)
{
	vector<int> rs(graph.size(), MAX_COST);
	bool *splitSet = new bool[graph.size()];
	for (unsigned i = 0; i < graph.size(); i++)
	{
		splitSet[i] = false;
	}
	rs[sp] = 0;

	for (unsigned i = 0; i < graph.size(); i++)
	{
		int id = minDist(rs, splitSet);
		if (-1 == id)break;//之后的都没路了

		splitSet[id] = true;
		for (unsigned j = 0; j < graph.size(); j++)
		{
			if (!splitSet[j] && -1 != graph[id][j])
			{
				int tmp = max(rs[id], graph[id][j]);
				rs[j] = min(rs[j], tmp);
			}
		}
	}
	delete [] splitSet;
	return rs;
}

void CountryRoads()
{
	int T = 0, n = 0, m = 0, u = 0, v = 0, w = 0, t = 0;
	cin>>T;
	for (int i = 0; i < T; i++)
	{
		cin>>n>>m;		
		vector<vector<int> > graph(n, vector<int>(n, -1));
		for (int j = 0; j < m; j++)
		{
			cin>>u>>v>>w;
			assert(0<=u && u<n && 0<=v && v<n);
			if (-1 == graph[u][v]) graph[u][v] = graph[v][u] = w;
			else graph[u][v] = graph[v][u] = min(graph[u][v], w);
		}
		cin>>t;
		vector<int> rs = findPaths(graph, t);
		cout<<"Case "<<i+1<<":"<<endl;
		for (unsigned i = 0; i < rs.size(); i++)
		{
			if (MAX_COST == rs[i]) cout<<"Impossible"<<endl;
			else cout<<rs[i]<<endl;
		}
	}
}
int main()
{
	CountryRoads();
	return 0;
}