HDU-1535 Invitation Cards

                                     Invitation Cards 

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. 

InputThe 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. 
OutputFor 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

英文题目emmmm，大概意思就是1到N个点最短路+N点到1最短路（有向图）

揣摩了一会题目，这个题只是需要一个反向建边然后 dijsktra两次就行

第一次 提交 数组开小了 RE 第二次 建立了两个邻接表 显然内存爆了 MLE 

后面想了下  我们先记录第一次dij的ans的值 然后重新清空邻接表 再反向建边  然后加上次的ans 输出 然后AC了

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<vector>
#include<stack>
#include<list>
#include<map>
using namespace std;
const int INF = 1e9;
const int maxn = 1000005;
typedef pair<int, int> pll;
struct node// 存输入数据的 结构体
{
    int u,v,w;
    //node(int nt,int wb):u(nt),v(wb) {};
} g[maxn];
struct no// 邻接表
{
    int a,b;
    no(int aa,int bb):a(aa),b(bb) {};
};
vector< no > p2[maxn];
int dis[maxn];
bool vis[maxn];
void init()// 清空 切记第一次dij之后需要 清空一次
{
    int i;
    for(i=0; i<maxn; i++)
    {
        p2[i].clear();
    }
}
void iit()
{
    int i;
    for(i=0; i<maxn; i++)
    {
        dis[i]=INF;
        vis[i]=0;
    }
}
int dijsktra( vector<no> p[],int n)
{
    priority_queue<pll, vector<pll>, greater<pll> >q;//greater 降序
    iit();
    pair<int,int > temp;
    q.push(pll(0,1));// pair first是 距离 second是点  
    dis[1]=0;
    while(!q.empty())
    {
        temp=q.top();
        q.pop();
        int x=temp.second;
        if(vis[x]) continue;
        vis[x]=1;
        for(int i=0; i<p[x].size(); i++)
        {
            int y=p[x][i].a;
            if(vis[y]) continue ;
            int d=p[x][i].b;
            if(dis[y]>dis[x]+d)
            {
                dis[y]=dis[x]+d;
                q.push(pll(dis[y],y));

            }
        }
    }
    int ans=0,i;
    for(i=1; i<=n; i++)
        ans+=dis[i];
    return ans;//返回一次值
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;

        scanf("%d %d",&n,&m);
        int i,j,a,b,c;
        for(i=0; i<m; i++)
        {
            scanf("%d %d %d",&g[i].u,&g[i].v,&g[i].w);
            p2[g[i].u].push_back(no(g[i].v,g[i].w));//开始没有定义 no 这个结构体 然后程序构建一直错（噗
        }
        int ans=0;
        ans=dijsktra(p2,n);//第一次dij
        init();//清空 邻接表
        for(i=0; i<m; i++)
            p2[g[i].v].push_back(no(g[i].u,g[i].w)); //反向建边
        ans+=dijsktra(p2,n);//累加
        cout<<ans<<endl;
        init();
    }
}