hdu 3996 Gold Mine 最大权闭合子图

Gold Mine

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2374    Accepted Submission(s): 514

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

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3996

 

题意：有一个矿，其中分为很多个区域。每个区域里面有黄金。每个黄金有价值，挖黄金需要一定的花费。但是挖其中的一些黄金必须要先挖另一些黄金。

 

思路：挖其中的一些黄金需要先挖另一些黄金。显示了必要条件。所以可以用最大权闭合子图来解决。

建图：增加源点s和汇点t。每个点和源点连一条边，容量为黄金的价值。每个点和汇点连一条边，容量为所需要的花费。

如果挖i个黄金需要先挖j黄金，那么从i到j连一条有向边，容量为正无穷。

要注意的是最后的结果可能会超int，要用long long。

#include <bits/stdc++.h>
using namespace std;
#define maxn 3200
const long long inf = 1ll<<60;
struct Edge
{
    int from, to;
    long long cap, flow;
    Edge(int f, int t, long long  c, long long  fl)
    {
        from = f; to = t; cap = c; flow = fl;
    }
};
long long min(long long a, long long b)
{
    return a<b?a:b;
};
vector <Edge> edges;
vector <int> G[maxn];
int n, m, s, t;
int cur[maxn], vis[maxn], d[maxn];
void AddEdge(int from, int to, long long cap)
{
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}
bool bfs()
{
    memset(vis, 0, sizeof(vis));
    d[s] = 0; 
    vis[s] = 1;
    queue <int> q;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int i = 0; i < G[u].size(); i++)
        {
            Edge &e = edges[G[u][i]];
            if(!vis[e.to] && e.cap > e.flow)
            {
                d[e.to] = d[u]+1;
                vis[e.to] = 1;
                q.push(e.to);
            }
        }
    }
    return vis[t];
}
long long dfs(int x, long long  a)
{    
    if(x == t || a == 0) return a;
    long long flow = 0, f;
    for(int &i = cur[x]; i < G[x].size(); i++)
    {
        Edge &e = edges[G[x][i]];
        if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edges[G[x][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}
long long Maxflow()
{
    long long flow = 0;
    while(bfs())
    {
        memset(cur, 0, sizeof(cur));
        flow += dfs(s, inf);
    }
    return flow;
}
int T, N;
int main()
{
    scanf("%d", &T);
    int cast = 0;
    while(T--)
    {
        cast++;
        edges.clear();
        for(int i = 0; i < maxn; i++) G[i].clear();
        scanf("%d", &N);
        s = 0; t = N*25+1; n = N;
        long long sum = 0;
        for(int i = 1; i <= N; i++)
        {
            int cnt; scanf("%d", &cnt);
            for(int j = 1; j <= cnt; j++)
            {
                long long value, cost;
                scanf("%lld%lld",&cost, &value);
                sum += value;
                AddEdge(s, (i-1)*25+j, value);
                AddEdge((i-1)*25+j, t, cost);
                int w; scanf("%d", &w);
                while(w--)
                {
                    int a, b; scanf("%d%d", &a, &b);
                    AddEdge((i-1)*25+j, (a-1)*25+b, inf);
                }
            }
        }
        long long flow = Maxflow();
        printf("Case #%d: %lld
", cast, sum-flow);
    }
    return 0;
}