【題目描述】

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

【題目分析】
求的是從源點到所有點以及所有點到源點的路程和最小，對于源點到所有點的路程和最小我們很自然的想到Dijkstra，因為數據規模比較大需要用鄰接表，我用的是前向星的方式進行儲存，雖然用vector也可以，但是聽說vector更有可能MLE（因為STL中vector實際要多用一倍的空間）和TLE（因為vector的一些挺簡單的操作耗時比一般數組要長）。
但是對于其他所有的點到源點的路程和最小我們直接用Dijkstra顯然是不行的，而用Floyed復雜度太高。
這個時候就需要一點點腦洞想到反向建圖（圖論中反向建圖經常會碰到，很多正面看起來難以解決的問題反過來看就會變的簡單起來），反向建圖后我們再跑一遍Dijkstra，這時候的路徑方向全部反過來就是實際上其他所有點到源點的路徑。然后再求和就可以啦。
前向星進行儲存的時候要注意head數組（保存每個點第一條邊的數組）如果初始化為0，邊一定要從1開始，否則無法判斷到底是沒有邊了還是指向第一個邊，如果習慣從0開始數邊應該將head數組初始化為-1.
將結構體作為優先隊列的元素的時候需要進行運算符重載，優先隊列默認是使用小于號進行比較的，我們應該對小于號進行重載。如果想要按照某個值從大到小重載后還是小于號，如果想要某個值從小到大應該重載為大于號，這是因為優先隊列默認是從大到小的，也就是“重載后值越小”的越在后面。
在對Dijkstra用優先隊列優化的時候不要進行初始化，也不要將源點設置為已經訪問過，如果設為已經訪問過就直接退出了。而將dis數組進行初始化以后我們后面就很有可能沒有辦法將其他節點放入隊列。
【AC代碼】

/*
*因為將兩個過程分別寫出來了顯得有點丑，空間也有點大
*我們也可以兩次建圖，用一套數組就好了
*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<climits>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;typedef long long ll;
const int MAXN=1e6+5;
struct node
{int v,next;ll w;node(int _v=0,int _next=0,ll _w=0):v(_v),w(_w),next(_next){}
}Edge1[MAXN],Edge2[MAXN];
struct qnode
{int x; ll d;qnode(int _x=0,ll _d=0):x(_x),d(_d){}bool operator <(const qnode &r)const {return d>r.d;}
}p;
int head1[MAXN],head2[MAXN],tot1,tot2;
bool vis[MAXN];
ll dis1[MAXN],dis2[MAXN],ans;
int n,m;void AddEdge1(int u,int v,ll w)
{tot1++;Edge1[tot1].v=v; Edge1[tot1].w=w; Edge1[tot1].next=head1[u];head1[u]=tot1;
}void AddEdge2(int u,int v,int w)
{tot2++;Edge2[tot2].v=v; Edge2[tot2].w=w; Edge2[tot2].next=head2[u];head2[u]=tot2;
}void Dijkstra1()
{memset(vis,0,sizeof(vis));memset(dis1,0x3f,sizeof(dis1));const int INF=dis1[1];int u,v;// for(int i=head1[1];i;i=Edge1[i].next)// {//     dis1[Edge1[i].v]=Edge1[i].w;// }dis1[1]=0;//vis[1]=true;priority_queue<qnode> q;q.push(qnode(1,0));while(!q.empty()){p=q.top(); q.pop();if(vis[p.x]) continue;vis[p.x]=true;u=p.x;for(int i=head1[u];i;i=Edge1[i].next){v=Edge1[i].v;if(!vis[v]  && dis1[v]>dis1[u]+Edge1[i].w){dis1[v]=dis1[u]+Edge1[i].w;q.push(qnode(v,dis1[v]));}}}
}void Dijkstra2()
{memset(vis,0,sizeof(vis));memset(dis2,0x3f,sizeof(dis2));const int INF=dis2[1];int u,v;// for(int i=head2[1];i;i=Edge2[i].next)// {//     dis2[Edge2[i].v]=Edge2[i].w;// }dis2[1]=0; //vis[1]=true;priority_queue<qnode> q;q.push(qnode(1,0));while(!q.empty()){p=q.top(); q.pop();if(vis[p.x]) continue;vis[p.x]=true;u=p.x;for(int i=head2[u];i;i=Edge2[i].next){v=Edge2[i].v;if(!vis[v] && dis2[v]>dis2[u]+Edge2[i].w){dis2[v]=dis2[u]+Edge2[i].w;q.push(qnode(v,dis2[v]));}}}
}void test()
{for(int i=1;i<=n;i++){printf("%lld ",dis1[i]);}printf("\n");for(int i=1;i<=n;i++){printf("%lld ",dis2[i]);}printf("\n");
}int main()
{int T,u,v; ll w;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);tot1=0; tot2=0;memset(head1,0,sizeof(head1));memset(head2,0,sizeof(head2));for(int i=0;i<m;i++){scanf("%d%d%lld",&u,&v,&w);AddEdge1(u,v,w); AddEdge2(v,u,w);}Dijkstra1(); Dijkstra2();//test();ans=0;for(int i=1;i<=n;i++){ans+=dis1[i]+dis2[i];}printf("%lld\n",ans);}return 0;
}