POJ2135 来回最短路（简单费用流）

题意：
      就是从1走到n然后再走回来，一条边只能走一次，要求路径最短。

思路：
      比较水，可以直接一遍费用流，不解释了，具体的看看代码，敲这个题就是为了练

练手，好久不敲了，怕比赛手生。

#include<queue>

#include<stdio.h>

#include<string.h>
#define N_node 1000 + 10

#define N_edge 40000 + 20

#define INF 100000000
using namespace std;
typedef struct

{

   int from ,to ,cost ,flow ,next;

}STAR;
STAR E[N_edge];

int list[N_node] ,tot;

int mer[N_edge];

int s_x[N_node];
void add(int a ,int b ,int c ,int d)

{

     E[++tot].from = a;

     E[tot].to = b;

     E[tot].cost = c;

     E[tot].flow = d;

     E[tot].next = list[a];

     list[a] = tot;

     

     E[++tot].from = b;

     E[tot].to = a;

     E[tot].cost = -c;

     E[tot].flow = 0;

     E[tot].next = list[b];

     list[b] = tot;

}
bool Spfa(int s ,int t ,int n)

{

    int mark[N_node] = {0};

    for(int i = 0 ;i <= n ;i ++) s_x[i] = INF;

    mark[s] = 1 ,s_x[s] = 0;

    queue<int>q;

    q.push(s);

    memset(mer ,255 ,sizeof(mer));

    while(!q.empty())

    {

        int xin ,tou = q.front();

        q.pop();

        mark[tou] = 0;

        for(int k = list[tou] ;k ;k = E[k].next)

        {

            xin = E[k].to;

            if(s_x[xin] > s_x[tou] + E[k].cost && E[k].flow)

            {

                s_x[xin] = s_x[tou] + E[k].cost;

                mer[xin] = k;

                if(!mark[xin])

                {

                    mark[xin] = 1;

                    q.push(xin);

                }

            }

        }

    }

    return mer[t] != -1;

}
int M_C_Flow(int s ,int t ,int n)

{

   int maxflow = 0 ,mincost = 0 ,minflow;

   while(Spfa(s ,t ,n))

   {

       minflow = INF;

       for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])

       if(minflow > E[i].flow) minflow = E[i].flow;

       for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])

       {

           E[i].flow -= minflow;

           E[i^1].flow += minflow;

           mincost += minflow * E[i].cost;

       }

       maxflow += minflow;

   }

   return mincost;

}
int main ()

{

    int n ,m ,i ,a ,b ,c;

    while(~scanf("%d %d" ,&n ,&m))

    {

       memset(list ,0 ,sizeof(list)) ,tot = 1;

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

       {

          scanf("%d %d %d" ,&a ,&b ,&c);

          add(a ,b ,c ,1);

          add(b ,a ,c ,1);

       }

       add(0 ,1 ,0 ,2);

       add(n ,n + 1 ,0 ,2);

       printf("%d\n" ,M_C_Flow(0 ,n + 1 ,n + 1));

    }

    return 0;

}