hdoj1385_Minimum Transport Cost（多源最短路）

最新推荐文章于 2017-08-20 09:31:59 发布

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最新推荐文章于 2017-08-20 09:31:59 发布

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分类专栏： acm_最短路径文章标签： acm 最短路 floyd

本文链接：https://blog.youkuaiyun.com/wdcxccsdn/article/details/46481565

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这篇博客探讨了如何运用Floyd算法解决HDOJ1385题目中的最小运输成本问题，通过详细解释算法流程，帮助读者理解多源最短路的求解策略。

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思路：floyd

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;

vector<int> pathcost[10000];
vector<int> pre[10000];
vector<int> cost[10000];
vector<int> citycost;

void floyd(int n)
{
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            {
                if (cost[i][j] > cost[i][k] + cost[k][j] + citycost[k])
                {
                    cost[i][j] = cost[i][k] + cost[k][j] + citycost[k];
                    pre[i][j] = pre[i][k];
                }
                else if (cost[i][j] == cost[i][k] + cost[k][j] + citycost[k])
                {
                    if (pre[i][j] > pre[i][k])
                    {
                        pre[i][j] = pre[i][k];
                    }
                }
            }
}

void solve(int s, int d)
{
    int inis = s, inid = d;
    s -= 1;
    d -= 1;
    vector<int> v;
    v.clear();
    v.push_back(s);
    while (pre[s][d]!=d)
    {
        v.push_back(pre[s][d]);
        s = pre[s][d];
    }
    v.push_back(d);

    cout << "From "<<inis<<" to "<<inid<<" :" << endl;
    cout << "Path: ";
    if (inis == inid)
        cout << inis;
    else{
        for (int i = 0; i < v.size(); i++)
        {
            cout << v[i] + 1;
            if (i < v.size() - 1)
                cout << "-->";
        }
    }
    cout << endl;
    cout << "Total cost : " <<cost[inis-1][inid-1] << endl;
}

int main()
{
    int n;
    while (cin >> n&&n)
    {
        for (int i = 0; i < n; i++)
        {
            pre[i].clear();
            pathcost[i].clear();
            cost[i].clear();
        }
        citycost.clear();
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            {
                int tpc;
                cin >> tpc;
                pathcost[i].push_back(tpc);
                if (pathcost[i][j] != -1)
                    cost[i].push_back(tpc);
                else
                    cost[i].push_back(99999999);
                pre[i].push_back(j);
            }
        for (int i = 0; i < n; i++)
        {
            int tcc;
            cin >> tcc;
            citycost.push_back(tcc);
        }
        floyd(n);
        int s, d;
        while (cin >> s >> d&&s != -1)
        {
            solve(s, d);
            cout << endl;
        }
    }
    return 0;
}