HDU 3499 Flight(dijkstra+思维）

最新推荐文章于 2018-09-04 15:14:42 发布

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本文介绍了一道关于寻找最经济飞行路线的算法题，通过Dijkstra算法计算最低成本，并考虑了特殊信用卡优惠条件下的最优解。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3499点击打开链接

Flight

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3203 Accepted Submission(s): 670

Problem Description

Recently, Shua Shua had a big quarrel with his GF. He is so upset that he decides to take a trip to some other city to avoid meeting her. He will travel only by air and he can go to any city if there exists a flight and it can help him reduce the total cost to the destination. There's a problem here: Shua Shua has a special credit card which can reduce half the price of a ticket ( i.e. 100 becomes 50, 99 becomes 49. The original and reduced price are both integers. ). But he can only use it once. He has no idea which flight he should choose to use the card to make the total cost least. Can you help him?

Input

There are no more than 10 test cases. Subsequent test cases are separated by a blank line.
The first line of each test case contains two integers N and M ( 2 <= N <= 100,000

0 <= M <= 500,000 ), representing the number of cities and flights. Each of the following M lines contains "X Y D" representing a flight from city X to city Y with ticket price D ( 1 <= D <= 100,000 ). Notice that not all of the cities will appear in the list! The last line contains "S E" representing the start and end city. X, Y, S, E are all strings consisting of at most 10 alphanumeric characters.

Output

One line for each test case the least money Shua Shua have to pay. If it's impossible for him to finish the trip, just output -1.

Sample Input

  
   4 4
Harbin Beijing 500
Harbin Shanghai 1000
Beijing Chengdu 600
Shanghai Chengdu 400
Harbin Chengdu

4 0
Harbin Chengdu

Sample Output

  
   800
-1



   
    
     Hint
    In the first sample, Shua Shua should use the card on the flight from
 Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the
 least total cost 800. In the second sample, there's no way for him to get to 
Chengdu from Harbin, so -1 is needed.

Author

Edelweiss

枚举所有边对每条边的起点计算由起点到该边的起点以及终点到该边的终点

#include<bits/stdc++.h>
using namespace std;
map<string ,int >mmap;
struct xjy
{
    int be;
    int en;
    long long int dis;
    bool operator < (const xjy &r)const
    {
        return dis>r.dis;
    }
};
vector<xjy> bes[100100],ens[100100];
int n,m;
long long int disbe[100100];
long long int disen[100100];
void bedijkstra(int be)
{
    for(int i=1;i<=n;i++)
        {
            disbe[i]=9223372036854775807;
        }
    priority_queue<xjy > q;
    xjy mid;
    mid.be=be;
    mid.dis=0;
    q.push(mid);
    disbe[be]=0;
    while(!q.empty())
    {
        mid=q.top();
        q.pop();
        for(int i=0;i<bes[mid.be].size();i++)
        {
            if((mid.dis+bes[mid.be][i].dis)<disbe[bes[mid.be][i].en])
            {
                disbe[bes[mid.be][i].en]=mid.dis+bes[mid.be][i].dis;
                xjy midmid;
                midmid.be=bes[mid.be][i].en;
                midmid.dis=disbe[bes[mid.be][i].en];
                q.push(midmid);
            }
        }
    }
}

void endijkstra(int be)
{
    for(int i=1;i<=n;i++)
        {
            disen[i]=9223372036854775807;
        }
    priority_queue<xjy > q;
    xjy mid;
    mid.be=be;
    mid.dis=0;
    q.push(mid);
    disen[be]=0;
    while(!q.empty())
    {
        mid=q.top();
        q.pop();
        for(int i=0;i<ens[mid.be].size();i++)
            {

                if(mid.dis+ens[mid.be][i].dis<disen[ens[mid.be][i].en])
                {
                    disen[ens[mid.be][i].en]=mid.dis+ens[mid.be][i].dis;
                    xjy midmid;
                    midmid.be=ens[mid.be][i].en;
                    midmid.dis=disen[ens[mid.be][i].en];
                    q.push(midmid);
                }

            }

    }
}
int main()
{
    while(cin >> n >> m)
    {
        mmap.clear();
        for(int i=1;i<100100;i++)
            bes[i].clear(),ens[i].clear();
        int cnt=1;
        for(int i=1;i<=m;i++)
        {
            long long int mid;
            string mid1,mid2;
            cin >> mid1>> mid2;
            cin >> mid;
            if(!mmap[mid1])
                mmap[mid1]=cnt++;
            if(!mmap[mid2])
                mmap[mid2]=cnt++;
            xjy midmid;
            midmid.dis=mid;
            midmid.be=mmap[mid1];
            midmid.en=mmap[mid2];
            bes[mmap[mid1]].push_back(midmid);
            midmid.dis=mid;
            midmid.be=mmap[mid2];
            midmid.en=mmap[mid1];
            ens[mmap[mid2]].push_back(midmid);
        }
        string mid1,mid2;
        cin >> mid1 >> mid2;
        if(!mmap[mid1]||!mmap[mid2])
            printf("-1\n");
        else
        {
            bedijkstra(mmap[mid1]);
            endijkstra(mmap[mid2]);
            long long int minn=9223372036854775807;
            for(int i=1;i<=n;i++)
                for(int j=0;j<bes[i].size();j++)
                {
                    minn=min(bes[i][j].dis/2+disbe[bes[i][j].be]+disen[bes[i][j].en],minn);
                }
            cout << minn << endl;
        }
    }
}