题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=3667
思路:平方关系,直接建图每次增广并不是最优。。。。
1^2=1,2^2=1+3,3^2=1+3+5,4^2=1+3+5+7.......
所以,对于每条边<x,y>,若流量为c,则在x与y之间连c条边,流量均为1,费用分别为a[i],3*a[i],5*a[i].........由于每次增广时流量相同时选择最小花费的边,若该边<x,y>流量为c,则总花费为a[x]+3*a[x]+5*a[x]+.........+(2*c-1)*a[x]=a[x]*c^2,符合题意。对新图求最小费用最大流即可,若最大流小于k,则无解。
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debu
using namespace std;
typedef long long LL;
const int maxn=200;
const int INF=0x3f3f3f3f;
struct Edge
{
int from,to,cap,flow,cost;
Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w) {}
};
queue<int> q;
vector<Edge> edges;
vector<int> G[maxn];
int n,m,k,p[maxn],a[maxn];
int inq[maxn],d[maxn];
LL cost;
void init(int n)
{
memset(p,0,sizeof(p));
memset(a,0,sizeof(a));
memset(d,0,sizeof(d));
for(int i=0; i<=n+1; i++) G[i].clear();
edges.clear();
}
void addedges(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));
int m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BF(int s,int t,int& flow,LL& cost)
{
while(!q.empty()) q.pop();
for(int i=0; i<=n; i++) d[i]=INF;
memset(inq,0,sizeof(inq));
d[s]=0,inq[s]=1,p[s]=0;
q.push(s);
a[s]=INF;
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 0;
flow+=a[t];
cost+=(LL)d[t]*(LL)(a[t]);
//cout<<a[t]<<" "<<d[t]<<" "<<flow<<" "<<cost<<endl;
for(int u=t; u!=s; u=edges[p[u]].from)
{
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
}
return 1;
}
int mincostmaxflow(int s,int t,LL& cost)
{
int flow=0;
cost=0;
while(BF(s,t,flow,cost));
return flow;
}
int main()
{
#ifdef debug
freopen("in.in","r",stdin);
#endif // debug
ios::sync_with_stdio(0);
while(cin>>n>>m>>k)
{
init(n);
for(int i=0; i<m; i++)
{
int x,y,c,w,tmp=1;
cin>>x>>y>>w>>c;
//cout<<x<<" "<<y<<" "<<w<<" "<<c<<endl;
for(int j=1; j<=c; j++)
{
addedges(x,y,1,w*tmp);
tmp+=2;
}
}
addedges(0,1,k,0);
//cout<<"123"<<endl;
int tmp=mincostmaxflow(0,n,cost);
//cout<<tmp<<endl;
if(tmp<k) cout<<-1<<endl;
else cout<<cost<<endl;
//cout<<"flag"<<endl;
}
return 0;
}