HDU3996 Gold Mine最大权闭合图 2011 Multi-University Training Contest 16 - Host by TJU

 

Gold Mine

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 225 Accepted Submission(s): 58

Problem Description

Long long ago, there is a gold mine.The mine consist of many layout, so some area is easy to dig, but some is very hard to dig.To dig one gold, we should cost some value and then gain some value. There are many area that have gold, because of the layout, if one people want to dig one gold in some layout, he must dig some gold on some layout that above this gold's layout. A gold seeker come here to dig gold.The question is how much value the gold he can dig, suppose he have infinite money in the begin.

Input

First line the case number.(<=10)

Then for every case:
one line for layout number.(<=100)
for every layout
first line gold number(<=25)
then one line for the dig cost and the gold value(32bit integer), the related gold number that must be digged first(<=50)

then w lines descripte the related gold followed, each line two number, one layout num, one for the order in that layout
see sample for details

Output

Case #x: y.
x for case number, count from 1.
y for the answer.

Sample Input

1
2
1
10 100 0
2
10 100 1
1 1
10 100 1
1 1

Sample Output

   
   
   
   

    
    
    
    Case #1: 270
   
   
   
   

Source

2011 Multi-University Training Contest 16 - Host by TJU

Recommend

lcy

 

和HDU3917Road constructions几乎一样， http://blog.csdn.net/power721/article/details/6665750

只是收益和支出合并成了一个点

 

建图：

s到金矿ij连边，边权为支出

金矿ij到t连边，边权为收益

要采金矿j必须先采金矿i，则i到j连边，边权为inf

 

代码：

#include<cstdio>   
#include<cstring>   
#include<vector>   
#define N 5005   
#define M 1000005   
#define inf 1ll<<60  
#define min(a,b) ((a)<(b)?(a):(b))   
using namespace std;  
  
__int64 n,m,s,t,num,adj[N],dis[N],q[N];  
struct edge  
{  
    __int64 v,w,pre;  
}e[M];  
void insert(__int64 u,__int64 v,__int64 w)  
{  
    e[num]=(edge){v,w,adj[u]};  
    adj[u]=num++;  
    e[num]=(edge){u,0,adj[v]};  
    adj[v]=num++;  
}  
__int64 bfs()  
{  
    __int64 i,x,v,head=0,tail=0;  
    memset(dis,0,sizeof(dis));  
    dis[s]=1;  
    q[++tail]=s;  
    while(head!=tail)  
    {  
        x=q[head=(head+1)%N];  
        for(i=adj[x];~i;i=e[i].pre)  
            if(e[i].w&&!dis[v=e[i].v])  
            {  
                dis[v]=dis[x]+1;  
                if(v==t)  
                    return 1;  
                q[tail=(tail+1)%N]=v;  
            }  
    }  
    return 0;  
}  
__int64 dfs(__int64 x,__int64 limit)  
{  
    if(x==t)  
        return limit;  
    __int64 i,v,tmp,cost=0;  
    for(i=adj[x];~i&&cost<limit;i=e[i].pre)  
        if(e[i].w&&dis[x]==dis[v=e[i].v]-1)  
        {  
            tmp=dfs(v,min(limit-cost,e[i].w));  
            if(tmp)  
            {  
                e[i].w-=tmp;  
                e[i^1].w+=tmp;  
                cost+=tmp;  
            }  
            else  
                dis[v]=-1;  
        }  
    return cost;  
}  
__int64 Dinic()  
{  
    __int64 ans=0;  
    while(bfs())  
        ans+=dfs(s,inf);  
    return  ans;  
}  
int main()  
{  
	__int64 c,cc=0;
	scanf("%I64d",&c);
    while(c--)  
    {  
        __int64 i,j,k,u,v,w1,w2,a,b,idx=2,sum=0,h[105][35]={0};  
        num=0;  
        memset(adj,-1,sizeof(adj));  
        s=0;  
        t=1; 
		scanf("%I64d",&n);
        for(i=0;i<n;i++)
		{
			scanf("%I64d",&m);
			for(j=0;j<m;j++)
			{
				scanf("%I64d%I64d%I64d",&w1,&w2,&k);
				sum+=w2;
				u=h[i][j]=idx++;
				insert(s,u,w1);
				insert(u,t,w2);
				while(k--)
				{
					scanf("%I64d%I64d",&a,&b);
					v=h[a-1][b-1];
					insert(v,u,inf);
				}
			}
		}
        printf("Case #%I64d: %I64d\n",++cc,sum-Dinic());  
    }  
}