POJ-1511: Invitation Cards（反向建边）

最短路

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

题意：输入t代表测试样例个数。有编号为1-n的n个城市，m条单向路，下面m行，三个数字a，b，c代表城市a到b之间距离为c。1是物资所在地，2-n是受灾地，每次只能救济一个城市，问救完所有城市又返回1所走的最短路。

分析：从1出发到编号为n的城市的最短路可用dij求得，从n出发回到1，就反向建边，正向遍历，也就是说，单项路是从n到1的，建立的时候存成1到n，那么找到每个dis加起来即为n到1的最短路。最后和正向的所有dis相加。套用dij模板即可。【然而姿势问题用我的vector写会超时，球球的竟然勉强能卡过QAQ
所以……以下是邻接表+优先队列优化版本的代码

• 代码：

#include<stdio.h>
#include<queue>
#include<string.h>
#define INF 0x7fffffff
const int maxn=1e6+5;
using namespace std;
int n,m;
int num,num2;
int vis[maxn],vis2[maxn];
int pos[maxn],pos2[maxn];
int dis[maxn],dis2[maxn];
struct node1
{
    int en,val,next;
}a[maxn],b[maxn];
struct node2
{
    int id,val; //id是位置
    node2(int id1=0,int val1=0) //构造函数
    {
        id=id1;val=val1;
    }
    friend bool operator <(node2 x,node2 y) //重载＜符号
    {
        return x.val>y.val; 
    }
};
void add(int st,int en,int val)
{
    a[num].en=en;
    a[num].val=val;
    a[num].next=pos[st];
    pos[st]=num;
    num++;
}
void add2(int st,int en,int val)
{
    b[num2].en=en;
    b[num2].val=val;
    b[num2].next=pos2[st];
    pos2[st]=num2;
    num2++;
}
void dij()
{
    dis[1]=0;
    node2 now;
    priority_queue<node2>q;
    q.push(node2(1,0));
    while(!q.empty())
    {
        int i;
        now=q.top();
        q.pop();
        if(vis[now.id]==1) continue;
        vis[now.id]=1;
        for(i=pos[now.id];i!=-1;i=a[i].next)
        {
            if(vis[a[i].en]==0&&dis[a[i].en]>dis[now.id]+a[i].val)
            {
                dis[a[i].en]=dis[now.id]+a[i].val;
                q.push(node2(a[i].en,dis[a[i].en]));
            }
        }
    }
}
void dij2()
{
    dis2[1]=0;
    node2 now;
    priority_queue<node2>q;
    q.push(node2(1,0));
    while(!q.empty())
    {
        int i;
        now=q.top();
        q.pop();
        if(vis2[now.id]==1) continue;
        vis2[now.id]=1;
        for(i=pos2[now.id];i!=-1;i=b[i].next)
        {
            if(vis2[b[i].en]==0&&dis2[b[i].en]>dis2[now.id]+b[i].val)
            {
                dis2[b[i].en]=dis2[now.id]+b[i].val;
                q.push(node2(b[i].en,dis2[b[i].en]));
            }
        }
    }
}
int main()
{
    int a,b,c;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        num=0,num2=0;
        for(int i=0;i<=n;i++)
        {
            dis[i]=dis2[i]=INF;
            vis[i]=vis2[i]=0;
            pos[i]=pos2[i]=-1;
        }
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add2(b,a,c);
        }
        dij();
        dij2();
        long long sum=0;
        for(int i=1;i<=n;i++)
            sum+=dis[i]+dis2[i];
        printf("%lld\n",sum);
    }
    return 0;
}