Open-Pit Mining

Description

Open-pit mining is a surface mining technique of extracting rock or minerals from the earth by their removal from an open pit or borrow. Open-pit mines are used when deposits of commercially useful minerals or rocks are found near the surface. Automatic Computer Mining (ACM) is a company that would like to maximize its profits by open-pit mining. ACM has hired you to write a program that will determine the maximum profit it can achieve given the description of a piece of land.

Each piece of land is modelled as a set of blocks of material. Block i has an associated value (vi), as well as a cost (ci), to dig that block from the land. Some blocks obstruct or bury other blocks. So for example if block i is obstructed by blocks j and k, then one must first dig up blocks j and k before block i can be dug up. A block can be dug up when it has no other blocks obstructing it.

Input

The first line of input is an integer N(1N200)N(1≤N≤200) which is the number of blocks. These blocks are numbered 1 through N. Then follow N lines describing these blocks. The ith such line describes block i and starts with two integers vi, ci denoting the value and cost of the ith block (0vi,ci200)(0≤vi,ci≤200) . Then a third integer 0miN10≤mi≤N−1 on this line describes the number of blocks that block i obstructs. Following that are mi distinct space separated integers between 1 and N (but excluding i) denoting the label(s) of the blocks that block i obstructs. You may assume that it is possible to dig up every block for some digging order. The sum of values mi over all blocks i will be at most 500.

Output

Output a single integer giving the maximum profit that ACM can achieve from the given piece of land.

Sample Input

5
0 3 2 2 3
1 3 2 4 5
4 8 1 4
5 3 0
9 2 0

Sample Output

2

Hint

数论:网络流:最大权闭合子图,模板题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>  
using namespace std;  
typedef long long ll;  
const int N=1e6+5;  
const int INF=0x3f3f3f3f;  
  
struct node{  
    ll t,cap,flow,next;    
}e[N];  
int head[N],cur[N],cnt; 
void init(){  
    memset(head,-1,sizeof(head));  
    cnt=0;  
}  
void add(int u,int v,ll cap)  
{  
    e[cnt]=node{v,cap,0,head[u]};  
    head[u]=cnt++;  
    e[cnt]=node{u,0,0,head[v]};    
    head[v]=cnt++;  
}  
int d[N];   
bool bfs(int s,int t)   //O(n+m)  
{  
    memset(d,0,sizeof(d));  
    queue<int>q;  
    q.push(s);  
    d[s]=1;  
    while(!q.empty())  
    {  
        int u=q.front();q.pop();  
        for(int i=head[u];~i;i=e[i].next)  
        {  
            int v=e[i].t;  
            if(d[v]==0&&e[i].cap-e[i].flow>0)  
            {  
                d[v]=d[u]+1;  
                q.push(v);  
            }  
        }  
    }  
    return d[t]>0;   
}  
ll dfs(int s,int t,ll minedge)  
{  
    if(s==t)return minedge;  
    ll flow=0;    
    for(int &i=cur[s];~i;i=e[i].next)  
    {  
        int v=e[i].t;  
        if(d[v]==d[s]+1&&e[i].cap-e[i].flow>0)  
        {  
            ll temp=dfs(v,t,min(minedge-flow,e[i].cap-e[i].flow));  
            e[i].flow+=temp;    
            e[i^1].flow-=temp;  
            flow+=temp;  
            if(flow==minedge)return flow;  
        }  
    }  
    if(flow==0)d[s]=0;    
    return flow;  
}  
ll dinic(int s,int t) 
{  
    ll maxflow=0;  
    while(bfs(s,t))     
    {  
        memcpy(cur,head,sizeof(head));   
        maxflow+=dfs(s,t,INF);  
    }  
    return maxflow;  
}  
int pro[220];  
int main()  
{  
    int n,u,v,x,k;  
    init();  
    scanf("%d",&n);  
    for(int i=1;i<=n;i++)  
    {  
        scanf("%d%d%d",&u,&v,&k);  
        pro[i]=u-v;  
        while(k--){  
            scanf("%d",&x);  
            add(x,i,INF);  
        }  
    }  
    ll sum=0;  
    for(int i=1;i<=n;i++)  
    {  
        if(pro[i]>0){  
            add(0,i,pro[i]);  
            sum+=pro[i];  
        }  
        else if(pro[i]<0)add(i,n+1,-pro[i]);  
    }  
    int ans=dinic(0,n+1);  
    cout<<sum-ans<<endl;  
    return 0;  
}  




posted @ 2018-05-27 22:14  StarHai  阅读(421)  评论(0编辑  收藏  举报