POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / SCU 1132 Invitation Cards / ZOJ 2008 Invitation Cards / HDU 1535 （图论，最短路径）

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

Http

POJ:https://vjudge.net/problem/POJ-1511

UVA:https://vjudge.net/problem/UVA-721

SPOJ:https://vjudge.net/problem/SPOJ-INCARDS

UVAlive:https://vjudge.net/problem/UVALive-5547

SCU:https://vjudge.net/problem/SCU-1132

ZOJ:https://vjudge.net/problem/ZOJ-2008

HDU:https://vjudge.net/problem/HDU-1535

Source

图论，最短路径

题目大意

派若干学生从1号点出发分别到达n个点，再回来。求所有学生走的最短路径

解决思路

思路很简单，把图正着反着各存一遍，分别对1号点跑最短路就可以了，最后累加起来的就是答案。

但是这题竟然卡vector，不能用vector存图，邻接矩阵也不行，必须用邻接表存。

为了方便，我把图和spfa都封装在结构体中了。

注意：SCU数据范围是50000，如果开1000000会爆空间

另外本题数据范围极大，需用读入优化+long long

代码

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<vector>

#include<queue>

using namespace std;

#define ll long long

const int maxN=1000001;

const int maxM=1000001;

const int inf=2147483647;

class Edge

{

public:

	int v;

	long long w;

};

class Graph//封装图的各种操作

{

public:

	int Head[maxN];//邻接表头指针

	ll Dist[maxN];//源点到各点的距离

	int Next[maxM];

	int V[maxM];//边的另一点

	ll W[maxM];//边权

private:

	int cnt;//建图时统计边数

	bool inqueue[maxN];//用于spfa

	queue<int> Q;

public:

	void init()//初始化

	{

		cnt=0;

		memset(Head,-1,sizeof(Head));

		memset(Next,-1,sizeof(Next));

		return;

	}

	void Add(int u,int v,int w)//加边

	{

		cnt++;

		V[cnt]=v;

		W[cnt]=w;

		Next[cnt]=Head[u];

		Head[u]=cnt;

		return;

	}

	void spfa(int S)//Spfa，源点是S

	{

		memset(Dist,127,sizeof(Dist));

		Dist[S]=0;

		memset(inqueue,0,sizeof(inqueue));

		while (!Q.empty())

		    Q.pop();

		Q.push(1);

		inqueue[1]=1;

		do

		{

			int u=Q.front();

			Q.pop();

			inqueue[u]=0;

			for (int i=Head[u];i!=-1;i=Next[i])

			{

				if (Dist[V[i]]>Dist[u]+W[i])

				{

					Dist[V[i]]=Dist[u]+W[i];

					if (inqueue[V[i]]==0)

					{

						Q.push(V[i]);

						inqueue[V[i]]=1;

					}

				}

			}

		}

		while (!Q.empty());

		return;

	}

};

int n,m;

Graph G1,G2;

int read();//读入优化

int main()

{

	int T=read();

	for (int ti=1;ti<=T;ti++)

	{

		n=read();m=read();

		G1.init();

		G2.init();

		for (int i=1;i<=m;i++)

		{

			int u=read(),v=read(),w=read();

			G1.Add(u,v,w);//正着存边

			G2.Add(v,u,w);//反着存边

		}

		G1.spfa(1);//分别跑spfa

		G2.spfa(1);

		ll Sum=0;

		for (int i=1;i<=n;i++)//统计答案

		    Sum+=G1.Dist[i]+G2.Dist[i];

		cout<<Sum<<endl;

	}

	return 0;

}

int read()

{

	int x=0;

	int k=1;

	char ch=getchar();

	while (((ch>'9')||(ch<'0'))&&(ch!='-'))

	    ch=getchar();

	if (ch=='-')

	{

		k=-1;

		ch=getchar();

	}

	while ((ch<='9')&&(ch>='0'))

	{

		x=x*10+ch-48;

		ch=getchar();

	}

	return x*k;

}