【题解】Luogu P2153 [SDOI2009]晨跑

原题传送门

一眼应该就能看出是费用流

因为每个交叉路口只能通过一次，所以我们进行拆点，连一条流量为1费用为0的边

再按照题目给的边（是单向边）建图

跑一下MCMF就行了

拆点很套路的~

#include <bits/stdc++.h>
#define N 205
#define M 20005
#define inf (1<<30)
#define getchar nc
using namespace std;
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
    register int x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void write(register int x)
{
    if(!x)putchar('0');if(x<0)x=-x,putchar('-');
    static int sta[20];register int tot=0;
    while(x)sta[tot++]=x%10,x/=10;
    while(tot)putchar(sta[--tot]+48);
}
inline int Min(register int a,register int b)
{
    return a<b?a:b;
}
struct node{
    int to,next,v,c;
}e[(M<<2)+(N<<1)];
int head[N<<1],cnt=1;
inline void add(register int u,register int v,register int val,register int cost)
{
    e[++cnt]=(node){v,head[u],val,cost};
    head[u]=cnt;
}
int n,m,s,t,maxflow=0,mincost=0;
int vis[N<<1],dist[N<<1];
inline bool spfa()
{
    memset(vis,0,sizeof(vis));
    memset(dist,0x3f,sizeof(dist));
    dist[s]=0;
    vis[s]=1;
    queue<int> q;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(register int i=head[u];i;i=e[i].next)
        {
            int v=e[i].to;
            if(dist[v]>dist[u]+e[i].c&&e[i].v)
            {
                dist[v]=dist[u]+e[i].c;
                if(vis[v]==0)
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    return dist[t]!=0x3f3f3f3f;
}
inline int dfs(register int u,register int flow)
{
    if(u==t)
    {
        vis[t]=1;
        maxflow+=flow;
        return flow;
    }
    int used=0;
    vis[u]=1;
    for(register int i=head[u];i;i=e[i].next)
    {
        int v=e[i].to;
        if((vis[v]==0||v==t)&&e[i].v&&dist[v]==dist[u]+e[i].c)
        {
            int tmp=dfs(v,Min(e[i].v,flow-used));
            if(tmp)
            {
                mincost+=e[i].c*tmp;
                used+=tmp;
                e[i].v-=tmp;
                e[i^1].v+=tmp;
            }
            if(used==flow)
                return used;
        }
    }
    return used;
}
inline void MCMF()
{
    while(spfa())
    {
        vis[t]=1;
        while(vis[t])
        {
            memset(vis,0,sizeof(vis));
            dfs(s,inf);
        }
    }
}
int main()
{
    n=read(),m=read();
    s=1,t=n;
    for(register int i=2;i<n;++i)
        add(i,i+n,1,0),add(i+n,i,0,0);
    for(register int i=1;i<=m;++i)
    {
        int u=read(),v=read(),cost=read();
        if(u==1)
            add(u,v,1,cost),add(v,u,0,-cost);
        else if(v==n)
            add(u+n,v,1,cost),add(v,u+n,0,-cost);
        else
            add(u+n,v,1,cost),add(v,u+n,0,-cost);
    }
    MCMF();
    write(maxflow),putchar(' '),write(mincost);
    return 0;
}