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 " ,M_C_Flow(0 ,n + 1 ,n + 1));
    }
    return 0;
}