UVA_10000_Longest Paths_uva 10000 longest paths-优快云博客

本文链接：https://blog.youkuaiyun.com/cxy7tv/article/details/51064836

本文介绍了C++中实现的Floyd-Warshall算法用于解决全加权图的最短路径问题，并通过DFS算法进行路径优化。通过实例演示了如何使用这些经典算法解决实际问题。

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#include<iostream>
#include<cstdio>
#include<sstream>
#include<string>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#pragma warning(disable:4996)
using std::cin;
using std::cout;
using std::endl;
using std::stringstream;
using std::string;
using std::vector;
using std::list;
using std::pair;
using std::set;
using std::multiset;
using std::map;
using std::multimap;
using std::stack;
using std::queue;
void floyd_warshall(vector<vector<int>>&graph)
{
	for (size_t k = 0; k < graph.size(); k++)
	{
		for (size_t i = 0; i < graph.size(); i++)
		{
			for (size_t j = 0; j < graph.size(); j++)
			{
				graph[i][j] = std::min(graph[i][j], graph[i][k] + graph[k][j]);
			}
		}
	}
}
pair<int,int> dfs(int vex,const multimap<int, int>&graph)
{
	bool flag = false;
	pair<int, int>maximum{ 0,-1 };
	for (auto iter = graph.begin(); iter != graph.end(); iter++)
	{
		if (iter->first == vex)
		{
			flag = true;
			auto tmp = dfs(iter->second, graph);
			if (maximum.second < tmp.second)
			{
				maximum = tmp;
			}
		}
	}
	if (flag)
	{
		return{ maximum.first,maximum.second + 1 };
	}
	return{vex,0};
}
int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int n,count=0;
    while(cin>>n)
    {
        if(!n)
        {
            break;
        }
		vector<vector<int>>graph(n, (vector<int>)n);
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				graph[i][j] = 1000000000;
			}
		}
        int initial;cin>>initial;initial--;
        int first,second;
        while(cin>>first>>second)
        {
            if(!first&&!second)
            {
                break;
            }
			graph[first - 1][second - 1] = -1;
        }
		floyd_warshall(graph);
		pair<int, int>result{ 0,0 };
		for (size_t i = 0; i < graph[initial].size(); i++)
		{
			if (graph[initial][i] < result.second)
			{
				result.first = i;
				result.second = graph[initial][i];
			}
		}
		printf("Case %d: The longest path from %d has length %d, finishing at %d.\n\n",++count,initial+1,-result.second,result.first+1);
    }
    return 0;
}