poj 1511 Invitation Cards

Invitation Cards

Time Limit: 8000MS		Memory Limit: 262144K
Total Submissions: 14080		Accepted: 4562

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

Source

Central Europe 1998

题意: 题意：在一个城市里，有n（1~1000000）个运输中心，有m条有向路连接着任意的两个运输中心。ACM组织（位于编号为1的运输中心）要派发p个雇佣员工早上前往这p个运输中心去发传单，晚上再让他们都回到组织中。问所有人所走路程的总和最短是多少？
思路：spfa最短路径+邻接表表示。先一次spfa求早上从顶点1到各点的最短路径和，再对其反向图，再用spfa求一次顶点1到各顶点的最短路径和，这个值其实就是晚上各点回到顶点1的最短路径和了。

思路：稠密矩阵采用邻接矩阵，稀疏矩阵采用邻接表

AC代码：

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
#define maxn 2000500
#define INF 1000000000
struct Node
{
  int cost;
 int next;
 int t;
}node[maxn];
int n,m;
int adj1[maxn],adj2[maxn];
int d[maxn],d1[maxn];
int vis[maxn],cnt[maxn];
int size;
void change( int x,int y,int w,int *adj)
{
  node[size].t=y;
  node[size].cost=w;
  node[size].next=adj[x];
  adj[x]=size++;
}
void spfa( int ss,int *dis,int adj[ ])
{
  queue<int>q;
  int i;
  memset(vis,0,sizeof(vis));
  memset(cnt,0,sizeof(cnt));
  for( i=1;i<=n;i++)
   dis[i]=INF;
   dis[ss]=0;
   q.push(ss);
   vis[ss]=1;
  while(!q.empty( ))
  {
 int tt=q.front( );
 q.pop( );
 vis[tt]=0;
 for( i=adj[tt];i!=-1;i=node[i].next)
 {
   int vv=node[i].t,ww=node[i].cost;
   if(dis[vv]-ww>dis[tt])
   {
   dis[vv]=ww+dis[tt];
   if(!vis[vv])
   {
   q.push(vv);
   vis[vv]=1;
   if(++cnt[vv]>n) return ;
   }
   }
    }
  }
  return ;
}
int main( )
{
 int test;
 int i,a,b,c;
 scanf("%d",&test);
 while(test--)
 {
 size=0;
   scanf("%d%d",&n,&m);
   for( i=1;i<=n;i++)
   {
  adj2[i]=-1,adj1[i]=-1;//adj[i]代表无边，0代表有一条边
   }
   for( i=1;i<=m;i++)
   {
 scanf("%d%d%d",&a,&b,&c);
 change(a,b,c,adj1);
 change(b,a,c,adj2);
   }
   spfa(1,d,adj1);
   spfa(1,d1,adj2);
   long long sum=0;
   for( i=1;i<=n;i++)
   {
 sum+=d[i]+d1[i];
   }
   printf("%lld\n",sum);
 }
 return 0;
}

链接： http://poj.org/problem?id=1511