poj 2135 Farm Tour 最小费用流 spfa优化 16_05

本文主要是介绍poj 2135 Farm Tour 最小费用流 spfa优化 16_05_14，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

http://poj.org/problem?id=2135

题意：给你n个节点，中间连接有m条边，每条边有一定的权值，求两种1号节点走到n号节点没有公共边的走法中

总的权值最小的走法，输出这个最小值；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
const int N = 500;
const int M=20000;
struct edge{
   int to,cap,cost,rev;
};
vector<edge> G[1005];
int dist[1005],inq[1005],prev[1005],prel[1005];
int n,m,x,y,c;
void add_edge(int u,int v,int cost)
{
    G[u].push_back(edge{v,1,cost,G[v].size()});
    G[v].push_back(edge{u,0,-cost,G[u].size()-1});
    //cout<<u<<" "<<G[u].size()<<endl;
}
int mincost(int s,int t,int f)
{
    int ans=0;
    while(f>0)
    {
      memset(dist,inf,sizeof(dist));
      memset(inq,0,sizeof(inq));
      dist[s]=0;
      queue<int> q;
      q.push(s);
      inq[s]=1;
      while(!q.empty())
      {
          int u=q.front();
          q.pop();inq[u]=0;
          for(int j=0;j<G[u].size();j++)
            {
                edge &e=G[u][j];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost)
                   {
                       dist[e.to]=dist[u]+e.cost;
                       prev[e.to]=u;
                       prel[e.to]=j;
                       if(!inq[e.to])
                       {
                           q.push(e.to);
                           inq[e.to]=1;
                       }
                   }
            }
       }
       for(int i=t;i>s;)
       {
           int f=prev[i];
           int j=prel[i];
           G[f][j].cap-=1;
           G[i][G[f][j].rev].cap+=1;
           ans+=G[f][j].cost;
           i=prev[i];
       }
       f-=1;
    }
    return ans;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&x,&y,&c);
            add_edge(x,y,c);
            add_edge(y,x,c);
        }
        printf("%d\n",mincost(1,n,2));
    }
    return 0;
}

上面是SPFA，下面是朴素的bellman

   #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
const int N = 500;
const int M=20000;
struct edge{
   int to,cap,cost,rev;
};
vector<edge> G[1005];
int dist[1005],inq[1005],prev[1005],prel[1005];
int n,m,x,y,c;
void add_edge(int u,int v,int cost)
{
    G[u].push_back(edge{v,1,cost,G[v].size()});
    G[v].push_back(edge{u,0,-cost,G[u].size()-1});
    //cout<<u<<" "<<G[u].size()<<endl;
}
int mincost(int s,int t,int f)
{
    int ans=0;
    while(f>0)
    {
      memset(dist,inf,sizeof(dist));
      dist[s]=0;
      bool update=true;
      while(update)
      {
        update=false;
        for(int u=1;u<n;u++)
          for(int j=0;j<G[u].size();j++)
            {
                edge &e=G[u][j];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost)
                   {
                       dist[e.to]=dist[u]+e.cost;
                       prev[e.to]=u;
                       prel[e.to]=j;
                       update=true;
                       //cout<<"4"<<endl;
                   }
            }
       }
       for(int i=t;i>s;)
       {
           int f=prev[i];
           int j=prel[i];
           G[f][j].cap-=1;
           G[i][G[f][j].rev].cap+=1;
           ans+=G[f][j].cost;
           i=prev[i];
       }
       f-=1;
    }
    return ans;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&x,&y,&c);
            add_edge(x,y,c);
            add_edge(y,x,c);
        }
        printf("%d\n",mincost(1,n,2));
    }
    return 0;
}

分析：两条路不能有任意一条公共边，就决定了这道题目只能用流量为2的最小费用流，而不是最短路

下面是wa的代码：注意边的连接：无向路

        #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
const int N = 500;
const int M=20000;
struct edge{
   int to,cap,cost,rev;
};
vector<edge> G[1005];
int dist[1005],inq[1005],prev[1005],prel[1005];
int n,m,x,y,c;
void add_edge(int u,int v,int cost)
{
    G[u].push_back(edge{v,1,cost,G[v].size()});
    G[v].push_back(edge{u,0,-cost,G[u].size()-1});
    //cout<<u<<" "<<G[u].size()<<endl;
}
int mincost(int s,int t,int f)
{
    int ans=0;
    while(f>0)
    {
      memset(dist,inf,sizeof(dist));
      dist[s]=0;
      bool update=true;
      while(update)
      {
        update=false;
        for(int u=1;u<n;u++)
          for(int j=0;j<G[u].size();j++)
            {
                edge &e=G[u][j];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost)
                   {
                       dist[e.to]=dist[u]+e.cost;
                       prev[e.to]=u;
                       prel[e.to]=j;
                       update=true;
                       //cout<<"4"<<endl;
                   }
            }
       }
       for(int i=t;i>s;)
       {
           int f=prev[i];
           int j=prel[i];
           G[f][j].cap-=1;
           G[i][G[f][j].rev].cap+=1;
           ans+=G[f][j].cost;
           i=prev[i];
       }
       f-=1;
    }
    return ans;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&x,&y,&c);
            add_edge(x,y,c);
            add_edge(y,x,c);
        }
        printf("%d\n",mincost(1,n,2));
    }
    return 0;
}