bzoj3876: [Ahoi2014&Jsoi2014]支线剧情

题意:给一幅图,从1开始,每条边有边权最少走一遍,可以在任意点退出,问最小花费

题解:上下界费用流,每个边都流一遍,然后为了保证流量平衡,新建源点汇点,跑费用流把流量平衡

/**************************************************************
    Problem: 3876
    User: walfy
    Language: C++
    Result: Accepted
    Time:140 ms
    Memory:2868 kb
****************************************************************/

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000009
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=500+10,maxn=100000+10,inf=0x3f3f3f3f;

bool vis[N];
int cnt,head[N],pre[N],path[N],dis[N],in[N],out[N];
struct edge{int to,Next,c,cost;}e[maxn];
void init(){memset(head,-1,sizeof head);cnt=0;}
void add(int u,int v,int c,int cost)
{
//    printf("%d %d %d %lld\n",u,v,c,cost);
    e[cnt].to=v;e[cnt].c=c;e[cnt].cost=cost;e[cnt].Next=head[u];head[u]=cnt++;
    e[cnt].to=u;e[cnt].c=0;e[cnt].cost=-cost;e[cnt].Next=head[v];head[v]=cnt++;
}
bool spfa(int s,int t)
{
    memset(pre,-1,sizeof pre);
    memset(dis,inf,sizeof dis);
    memset(vis,0,sizeof vis);
    dis[s]=0;vis[s]=1;
    queue<int>q;q.push(s);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        vis[x]=0;
        for(int i=head[x];~i;i=e[i].Next)
        {
            int te=e[i].to;
            if(e[i].c>0&&dis[x]+e[i].cost<dis[te])
            {
                dis[te]=dis[x]+e[i].cost;
                pre[te]=x;path[te]=i;
                if(!vis[te])q.push(te),vis[te]=1;
            }
        }
    }
    return pre[t]!=-1;
}
int mincostmaxflow(int s,int t)
{
    int cost=0,flow=0;
    while(spfa(s,t))
    {
        int f=inf;
        for(int i=t;i!=s;i=pre[i])
            if(e[path[i]].c<f)
               f=e[path[i]].c;
        flow+=f;cost+=dis[t]*f;
        for(int i=t;i!=s;i=pre[i])
        {
            e[path[i]].c-=f;
            e[path[i]^1].c+=f;
        }
    }
    return cost;
}
int main()
{
    init();
    int n,tot=0;scanf("%d",&n);
    int s=1,t=n+1,ss=n+2,tt=n+3,ans=0;
    for(int i=1,k;i<=n;i++)
    {
        for(scanf("%d",&k);k;k--)
        {
            int a,b;scanf("%d%d",&a,&b);
            add(i,a,5000,b);
            out[i]++,in[a]++;ans+=b;
        }
        add(i,t,5000,0);
    }
    add(t,s,5000,0);
    for(int i=1;i<=n;i++)
    {
        if(in[i]>out[i])add(ss,i,in[i]-out[i],0);
        else add(i,tt,out[i]-in[i],0);
    }
    ans+=mincostmaxflow(ss,tt);
    printf("%d\n",ans);
    return 0;
}
/********************
6
2 2 1 3 2
2 4 3 5 4
2 5 5 6 6
0
0
0
********************/