【[SDOI2009]晨跑】

板子

题意就是每个点只能经过一次

所以非常显然拆点，除去(1,n)每个点(i)向(i')连一条容量为(1)费用为(0)的边

剩下的边按照输入给出的建就好了

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define maxn 405
#define re register
#define LL long long
#define inf 999999999
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
	char c=getchar();int x=0;while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();
	return x;
}
struct E{int v,nxt,f,w;}e[60005];
int head[maxn],d[maxn],vis[maxn];
int n,m,num=1,S,T;
inline void add(int x,int y,int ca,int z){e[++num].v=y;e[num].nxt=head[x],e[num].w=z;e[num].f=ca;head[x]=num;}
inline void C(int x,int y,int ca,int z){add(x,y,ca,z),add(y,x,0,-z);}
inline int SPFA()
{
	std::queue<int> q;
	for(re int i=S;i<=T;i++) d[i]=inf,vis[i]=0;
	q.push(T),vis[T]=0,d[T]=0;
	while(!q.empty())
	{
		int k=q.front();q.pop();vis[k]=0;
		for(re int i=head[k];i;i=e[i].nxt)
		if(e[i^1].f&&d[e[i].v]>d[k]+e[i^1].w)
		{
			d[e[i].v]=d[k]+e[i^1].w;
			if(!vis[e[i].v]) q.push(e[i].v),vis[e[i].v]=1;
		}
	}
	return d[S]<inf;
}
int dfs(int x,int now)
{
	if(x==T||!now) return now;
	int flow=0,ff;vis[x]=1;
	for(re int i=head[x];i;i=e[i].nxt)
	if(e[i].f&&!vis[e[i].v]&&d[x]+e[i^1].w==d[e[i].v])
	{
		ff=dfs(e[i].v,min(e[i].f,now));
		if(ff<=0) continue;
		now-=ff,flow+=ff;
		e[i].f-=ff,e[i^1].f+=ff;
		if(!now) break;
	}
	return flow;
}
int main()
{
	n=read(),m=read();C(1,n+1,inf,0),C(n,n+n,inf,0);
	for(re int i=2;i<n;i++) C(i,i+n,1,0);
	S=1,T=n+n;
	int x,y,z;
	for(re int i=1;i<=m;i++) x=read(),y=read(),z=read(),C(x+n,y,1,z);
	int tot=0,ans=0;
	while(SPFA())
	{
		vis[T]=1;
		while(vis[T])
		{
			for(re int i=S;i<=T;i++) vis[i]=0;
			int F=dfs(S,inf);
			tot+=F,ans+=F*d[S];
		}
	}
	printf("%d %d
",tot,ans);
	return 0;
}