bzoj

sol

每天每座城市建一個點，該怎麽連邊就怎麽連邊，每次只需要在上一次的殘余網絡上跑最大流，所以復雜度不會太高。

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;
}

[BZOJ1570][JSOI2008]Blue Mary的旅行