洛谷$P2604\ [ZJOI2010]$网络扩容 网络流

正解:网络流

解题报告:

传送门$QwQ$

昂第一问跑个最大流就成不说$QwQ$

然后第二问,首先原来剩下的边就成了费用为0的边?然后原来的所有边连接的两点都给加上流量为$inf$费用为$w$的边,保证流量等于$k$就给$S$向一号节点连流量为$k$费用为0的边就好,,,

感$jio$这题有点儿简单,,,所以为什么是紫昂洛谷评分好迷昂$QwQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define t(i) edge[i].to
#define n(i) edge[i].nxt
#define w(i) edge[i].wei
#define fy(i) edge[i].fy
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define e(i,x) for(ri i=head[x];~i;i=n(i))

const int N=+,M=+,inf=1e9;
int head[N],ed_cnt=-,S,T,dep[N],cur[N],dis[N],fr_nod[N],fr_ed[N],cost,n,m,K,x[M],y[M],w[M],f[M];
bool vis[N];
struct ed{int to,nxt,wei,fy;}edge[M<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void ad(ri x,ri y,ri z,ri p)
{edge[++ed_cnt]=(ed){x,head[y],z,p};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],,-p};head[x]=ed_cnt;}
il bool bfs()
{
    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;
    while(!Q.empty()){ri nw=Q.front();Q.pop();e(i,nw)if(w(i) && !dep[t(i)])dep[t(i)]=dep[nw]+,Q.push(t(i));}
    return dep[T];
}
il int dfs(ri nw,ri flow)
{
    if(nw==T || !flow)return flow;ri ret=;
    for(ri &i=cur[nw];~i;i=n(i))
        if(w(i) && dep[t(i)]==dep[nw]+)
        {ri tmp=dfs(t(i),min(flow,w(i)));flow-=tmp,w(i)-=tmp,w(i^)+=tmp,ret+=tmp;}
    return ret;
}
il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}
il bool spfa()
{
    queue<int>Q;Q.push(S);memset(dis,,sizeof(dis));dis[S]=;vis[S]=;
    while(!Q.empty())
    {
        ri nw=Q.front();Q.pop();vis[nw]=;
        e(i,nw)
        {
            if(w(i) && fy(i)+dis[nw]<dis[t(i)])
            {dis[t(i)]=dis[nw]+fy(i);fr_nod[t(i)]=nw;fr_ed[t(i)]=i;if(!vis[t(i)])Q.push(t(i)),vis[t(i)]=;}
        }
    }
    if(dis[T]==dis[T+])return ;
    ri flow=dis[T+];
    for(ri i=T;i!=S;i=fr_nod[i])flow=min(flow,w(fr_ed[i]));
    for(ri i=T;i!=S;i=fr_nod[i])w(fr_ed[i])-=flow,w(fr_ed[i]^)+=flow;
    cost+=flow*dis[T];return ;
}

int main()
{
    freopen("2604.in","r",stdin);freopen("2604.out","w",stdout);
    memset(head,-,sizeof(head));n=read();m=read();K=read();S=;T=n;
    rp(i,,m){x[i]=read(),y[i]=read(),w[i]=read(),f[i]=read();ad(y[i],x[i],w[i],);}
    printf("%d ",dinic());S=;rp(i,,m)ad(y[i],x[i],K,f[i]);ad(,S,K,);
    while(spfa());
    printf("%d\n",cost);
    return ;
}