[BZOJ1570][JSOI2008]Blue Mary的旅行

bzoj

sol

每天每座城市建一个点，该怎么连边就怎么连边，每次只需要在上一次的残余网络上跑最大流，所以复杂度不会太高。

code

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int gi()
{
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
const int N = 1e5+5;
const int inf = 1e9;
struct edge{int to,nxt,w;}a[N<<5];
int n,m,num,uu[N],vv[N],ww[N],id[51][2018],tot,S,T,head[N],cnt=1,dep[N],cur[N],flow,ans;
queue<int>Q;
void link(int u,int v,int w)
{
	a[++cnt]=(edge){v,head[u],w};
	head[u]=cnt;
	a[++cnt]=(edge){u,head[v],0};
	head[v]=cnt;
}
bool bfs()
{
	memset(dep,0,sizeof(dep));
	dep[S]=1;Q.push(S);
	while (!Q.empty())
	{
		int u=Q.front();Q.pop();
		for (int e=head[u];e;e=a[e].nxt)
			if (a[e].w&&!dep[a[e].to])
				dep[a[e].to]=dep[u]+1,Q.push(a[e].to);
	}
	return dep[T];
}
int dfs(int u,int f)
{
	if (u==T) return f;
	for (int &e=cur[u];e;e=a[e].nxt)
		if (a[e].w&&dep[a[e].to]==dep[u]+1)
		{
			int tmp=dfs(a[e].to,min(a[e].w,f));
			if (tmp) {a[e].w-=tmp;a[e^1].w+=tmp;return tmp;}
		}
	return 0;
}
int Dinic()
{
	int res=0;
	while (bfs())
	{
		for (int i=1;i<=tot;++i) cur[i]=head[i];
		while (int tmp=dfs(S,inf)) res+=tmp;
	}
	return res;
}
int main()
{
	n=gi();m=gi();num=gi();
	for (int i=1;i<=m;++i) uu[i]=gi(),vv[i]=gi(),ww[i]=gi();
	S=++tot;T=++tot;
	for (int i=1;i<=n;++i) id[i][0]=++tot;
	link(S,id[1][0],num);
	while (++ans)
	{
		for (int i=1;i<=n;++i) id[i][ans]=++tot,link(id[i][ans-1],id[i][ans],inf);
		for (int i=1;i<=m;++i) link(id[uu[i]][ans-1],id[vv[i]][ans],ww[i]);
		link(id[n][ans],T,num);flow+=Dinic();if (flow==num) break;
	}
	printf("%d\n",ans);return 0;
}