NWERC 2012 Problem A Admiral

一个最小费用最大流的简单建模题；

比赛的时候和小珺合力想到了这个题目的模型；

方法：拆点+边的容量为1

这样就可以保证他们不会在点上和边上相遇了！

感谢刘汝佳大神的模板，让我这个网络流的小白A了这个题。

代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define maxn 42005
#define inf 99999
using namespace std;

struct edge
{
int from ,to,cap,flow,cost;
};

struct mcmf
{
int n,m,s,t;
vector<edge>edges;
vector<int>g[maxn];
int inq[maxn],d[maxn],p[maxn],a[maxn];
void init(int n)
{
this->n=n;
edges.clear();
for(int i=0; i<n; i++)g[i].clear();
}
void addedge(int from,int to,int cap,int cost)
{
edges.push_back((edge){from,to,cap,0,cost});
edges.push_back((edge){to,from,0,0,-cost});
m=edges.size();
g[from].push_back(m-2);
g[to].push_back(m-1);
}
bool bellmamford(int s,int t,int &flow,int& cost)
{
for(int i=0; i<n; i++)d[i]=inf;
memset(inq,0,sizeof inq);
d[s]=0;
inq[s]=1;
p[s]=0;
a[s]=inf;
queue<int>q;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
inq[u]=0;
for(int i=0; i<g[u].size(); i++)
{
edge& e=edges[g[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost)
{
d[e.to]=d[u]+e.cost;
p[e.to]=g[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to])
{
q.push(e.to);
inq[e.to]=1;
}
}
}
}
if(d[t]==inf)return false;
flow+=a[t];
cost+=d[t]*a[t];
int u=t;
while(u!=s)
{
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return true;
}

int mincost(int s,int t)
{
int flow=0,cost=0;
while(bellmamford(s,t,flow,cost));
return cost;
}
}getans;

int main()
{
int nn,mm,f,t,c;
while(scanf("%d%d",&nn,&mm)!=EOF)
{
getans.init(2*nn+2);
getans.addedge(0,1+nn,2,0);
getans.addedge(nn,2*nn+1,2,0);
for(int i=2;i<nn;i++)
getans.addedge(i,i+nn,1,0);
for(int i=0; i<mm; i++)
{
scanf("%d%d%d",&f,&t,&c);
getans.addedge(f+nn,t,1,c);
}
printf("%d\n",getans.mincost(0,2*nn+1));
}
return 0;
}

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