SPFA最短路建图模板 POJ1511&&POJ3159

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本文介绍了一种利用最短路径算法来最小化每日交通费用的方法，帮助Antique Comedians of Malidinesia剧团合理分配学生志愿者到各个公交站进行宣传工作。

J - Invitation Cards

Time Limit:8000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

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

Sample Output

46
210



需要建二次图：都是单源最短路；
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
//const int inf=1e10;
#define inf 0xFFFFFFFF
typedef struct
{
    int w,v,next;
} node;
node edge[1000010];
int a[1000010][3];

int edgehead[1000010];
long long dis[1000010];
bool vis[1000010];
int n,m;int k;
void jia(int u,int v,int w)
{
    edge[k].v=v;
    edge[k].w=w;
    edge[k].next=edgehead[u];
    edgehead[u]=k++;
}
long long spfa()
{
    memset(vis,0,sizeof(vis));
    for(int i=2; i<=n; i++)
    {
        dis[i]=inf;
    }
    dis[1]=0;
    queue<int>q;
    q.push(1);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=false;
        for(int i=edgehead[u]; i; i=edge[i].next)
        {
            int v=edge[i].v;
            int w=edge[i].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=true;
                }
            }
        }
    }
    long long ans=0;
    for(int i=1; i<=n; i++)
        ans+=dis[i];
    return ans;
}
int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {

        
        memset(edgehead,0,sizeof(edgehead));
        memset(edge,0,sizeof(edge));
        k=1;scanf("%d%d",&n,&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);
            jia(a[i][0],a[i][1],a[i][2]);
        }
        long long ans=spfa();
        memset(edgehead,0,sizeof(edgehead));
        memset(edge,0,sizeof(edge));
        k=1;
        for(int i=1; i<=m; i++)
        {

            jia(a[i][1],a[i][0],a[i][2]);
        }
        ans+=spfa();
        printf("%lld\n",ans);
    }
}






    
     
      K - Candies
     
     Time Limit:1500MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u
     
      Submit 
      Status
     
    
    
     
      Description
      
       
        
        During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.
        snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?
       
      
     
     
      Input
      
       
        The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.
       
      
     
     
      Output
      
       
        Output one line with only the largest difference desired. The difference is guaranteed to be finite.
       
      
     
     
      Sample Input
      
       2 2
1 2 5
2 1 4
      
     
     
      Sample Output
      
       5 
       #include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int inf=0x3f3f3f3f;
int edgehead[30010];
int vis[30010];
int q[30010];
int dis[30010];
typedef struct
{
    int w,v,next;
} node;
node edge[150010];
int tol;
void jia(int u,int v,int w)//加边
{
    edge[tol].w=w;
    edge[tol].v=v;
    edge[tol].next=edgehead[u];
    edgehead[u]=tol++;
}
void  spfa(int start,int n)
{
    int top=0;
    for(int i=1; i<=n; i++)
    {
        if(i==start)
        {
            q[top++]=i;
            vis[i]=true;
            dis[i]=0;
        }
        else
        {
            vis[i]=false;
            dis[i]=inf;
        }
    }
    while(top!=0)
    {
        int u=q[--top];
        vis[u]=false;
        for(int i=edgehead[u]; i; i=edge[i].next)
        {
            int v=edge[i].v;


            if(dis[v]>dis[u]+edge[i].w)
            {
                dis[v]=dis[u]+edge[i].w;
                if(!vis[v])
                {
                    vis[v]=true;
                    q[top++]=v;
                }
            }
        }
    }
}
int main()
{
    int n,m;
    int a,b,c;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        tol=1;
        memset(edgehead,0,sizeof(edgehead));
        while(m--)
        {

            scanf("%d%d%d",&a,&b,&c);
            jia(a,b,c);
        }
        spfa(1,n);
        printf("%d\n",dis[n]);
    }
}



      
     
     
      

      

      

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