HDU 1535 Invitation Cards (最短路)

题目链接

Problem Description

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

Sample Input

2

2 2

1 2 13

2 1 33

4 6

1 2 10

2 1 60

1 3 20

3 4 10

2 4 5

4 1 50

Sample Output

46

210

分析：

从1号车站出发，要到达其余的n-1个车站并且回来，求所有路径的最小值。

首先明白一点我们求最短路的时候，是可以求一个点到其余的所有的点的，所以回来的时候我们直接从哪个点往回求有点困难，所有我们可以把所有的路径反向存储一下，那么这样的话反向后的从1出发的路径，就相当于原来的从其他点到达1的路径。正向走一次，反向走一次，就行了 。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
typedef long long LL;
#define INF 0xFFFFFFFF
const int maxn = 1000010;
struct edge
{
    int v,w,next;
} edges[2][maxn];

int head[2][maxn];
int e0,e1,n,m;
LL ans;
bool vis[maxn];
LL d[maxn];

void addedges(int u,int v,int w)//正向和反向都存一下路径
{
    edges[0][e0].v = v;
    edges[0][e0].w = w;
    edges[0][e0].next = head[0][u];
    head[0][u] = e0++;
    edges[1][e1].v = u;
    edges[1][e1].w = w;
    edges[1][e1].next = head[1][v];
    head[1][v] = e1++;
}

void spfa(int type)
{
    queue<int> q;
    for(int i=1; i<=n; i++)
    {
        d[i] = INF;
        vis[i] = 0;
    }
    d[1] = 0;
    q.push(1);
    vis[1] = 1;

    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i=head[type][u]; i!=-1; i=edges[type][i].next)
        {
            int v = edges[type][i].v;
            if(d[v] > d[u] + edges[type][i].w)
            {
                d[v] = d[u] + edges[type][i].w;
                if(vis[v] == 0)
                {
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}

int main()
{
    int t,u,v,w;
    scanf("%d",&t);
    while(t--)
    {
        memset(head,-1,sizeof(head));
        ans = e0 = e1 = 0;
        scanf("%d%d",&n,&m);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedges(u,v,w);
        }
        spfa(0);
        for(int i=1; i<=n; i++) ans += d[i];
        spfa(1);
        for(int i=1; i<=n; i++) ans += d[i];
        printf("%I64d\n",ans);
    }
    return 0;
}