需要注意的就是费用为a*x^2,这里的1<=x<=5
所以一条边可以拆成x条流量为1的边。
增加一条边花费为(2*x-1).
附上代码喵:
#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
using namespace std;
const int maxn=105;
const int maxm=100010;
#define INF 1000000000
#define MIN(a,b) a<b?a:b
int n,m,k;
int src,des;
struct EDGE
{
int u,v,w,c,next;
}edge[maxm];
int head[4*maxn],pre[4*maxn];
bool vis[4*maxn];int dist[4*maxn],delta[4*maxn];
int cnt;
void addedge(int u,int v,int w,int c)
{
edge[cnt].u=u;edge[cnt].v=v;edge[cnt].w=w;edge[cnt].c=c;edge[cnt].next=head[u];head[u]=cnt++;
edge[cnt].u=v;edge[cnt].v=u;edge[cnt].w=0;edge[cnt].c=-c;edge[cnt].next=head[v];head[v]=cnt++;
}
void buildmap()
{
src=0;des=n;cnt=0;
int u,v,a,w;
memset(head,-1,sizeof(head));
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d",&u,&v,&a,&w);
for(int j=1;j<=w;j++)
{
addedge(u,v,1,a*(2*j-1));
}
}
addedge(src,1,k,0);
}
bool SPFA()
{
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
for(int i=0;i<=des;i++)dist[i]=INF;
queue<int>q;
q.push(src);
vis[src]=true;dist[src]=0;delta[src]=INF;
while(!q.empty())
{
int k=q.front();q.pop();vis[k]=false;
for(int i=head[k];i!=-1;i=edge[i].next)
{
int u=edge[i].v;
if(edge[i].w>0&&dist[u]>dist[k]+edge[i].c){
dist[u]=dist[k]+edge[i].c;
pre[u]=i;
delta[u]=MIN(delta[k],edge[i].w);
if(!vis[u])
{
vis[u]=true;
q.push(u);
}
}
}
}
return pre[des]!=-1;
}
void update()
{
int x=des;
while(x!=src)
{
edge[pre[x]].w-=delta[des];
edge[pre[x]^1].w+=delta[des];
x=edge[pre[x]].u;
}
}
void MCMF()
{
buildmap();
int flow=0,cost=0;
while(SPFA()){
cost+=dist[des];
flow+=delta[des];
update();
}
if(flow<k)printf("-1\n");
else
printf("%d\n",cost);
}
int main()
{
while(scanf("%d%d%d",&n,&m,&k)!=EOF){
MCMF();
}
return 0;
}