HDU - 4780费用流

题意:M台机器要生产n个糖果，糖果i的生产区间在(si, ti)，花费是k(pi-si)，pi是实际开始生产的时间机器,j从初始化到生产糖果i所需的时间Cij，花费是Dij,任意机器从生产糖果i到生产糖果j，需花费时间Eij,花费Fij求生产完所有糖果所需的最小时间？

题解:先把糖果拆点,s向i糖果连流量为1,费用为0,i+n糖果向t连流量为1,费用为0,n2+i机器向t连流量为1,费用为0的边,如果i能流向j,那么j连i+n一条边流量为1,费用为(ti-sj)k+dij,如果机器j能流向i+2n,那么连一条边流量为1,费用为(p-s)*k+d,假设i流向了j,那么说明j在i之后生产,而且保证了每台机器只生产一种糖果

//#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 1000000007

#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 cd complex<double>

#define ull unsigned long long

#define base 1000000000000000000

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

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 void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

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 double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=300+10,maxn=400000+10,inf=0x3f3f3f3f;

struct edge{

    int to,Next,c;

    int cost;

}e[maxn];

int cnt,head[N];

int s,t;

int dis[N],pre[N],path[N];

bool vis[N];

void add(int u,int v,int c,int 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()

{

    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 n,m,k;

int mincostmaxflow()

{

    int cost=0,flow=0;

    while(spfa())

    {

        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;

        }

    }

    if(flow!=n)return -1;

    return cost;

}

void init()

{

    memset(head,-1,sizeof head);

    cnt=0;

}

int cans[N],cant[N],ti[N][N];

int main()

{

    while(~scanf("%d%d%d",&n,&m,&k))

    {

        if(!n&&!m&&!k)break;

        init();

        s=2*n+m+1,t=2*n+m+2;

        for(int i=1;i<=m;i++)add(2*n+i,t,1,0);

        for(int i=1;i<=n;i++)

        {

            scanf("%d%d",&cans[i],&cant[i]);

            add(s,i,1,0);add(i+n,t,1,0);

        }

        for(int i=1;i<=n;i++)

            for(int j=1;j<=m;j++)

                scanf("%d",&ti[i][j]);

        for(int i=1;i<=n;i++)

        {

            for(int j=1,x;j<=m;j++)

            {

                scanf("%d",&x);

                if(ti[i][j]<cant[i])

                {

                    add(i,2*n+j,1,max(0,ti[i][j] - cans[i])*k+x);

                }

            }

        }

        for(int i=1;i<=n;i++)

            for(int j=1;j<=n;j++)

                scanf("%d",&ti[i][j]);

        for(int i=1;i<=n;i++)

        {

            for(int j=1,x;j<=n;j++)

            {

                scanf("%d",&x);

                if(cant[i] + ti[i][j] < cant[j] && i != j)add(j,n+i,1,max(0,cant[i] + ti[i][j] - cans[j])*k+x);

            }

        }

        printf("%d\n",mincostmaxflow());

    }

    return 0;

}

/********************

3 2 1

4 7

2 4

8 9

4 4

3 3

3 3

2 8

12 3

14 6

-1 1 1

1 -1 1

1 1 -1

-1 5 5

5 -1 5

5 5 -1

********************/

巴特西

HDU - 4780费用流

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