题目链接：http://poj.org/problem?id=3469
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Dual Core CPU
Description As more and more computers are equipped with dual core CPU, SetagLilb, the Chief Technology Officer of TinySoft Corporation, decided to update their famous product - SWODNIW. The routine consists of N modules, and each of them should run in a certain core. The costs for all the routines to execute on two cores has been estimated. Let‘s define them as Ai and Bi. Meanwhile, M pairs Input There are two integers in the first line of input data, N and M (1 ≤ N ≤ 20000, 1 ≤ M ≤ 200000) . The next N lines, each contains two integer, Ai and Bi. In the following M lines, each contains three integers: a, b, w. The meaning is that if module a and module b don‘t execute on the same core, you should pay extra w dollars for the data-exchange Output Output only one integer, the minimum total cost. Sample Input 3 1 1 10 2 10 10 3 2 3 1000 Sample Output 13 Source POJ Monthly--2007.11.25, Zhou Dong |
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一道典型的最小割问题。
我们先假定两个额外的点，源点s，和汇点t。 我们记图中 s-t 割所对应的包含s的顶点集合为S,包含t集合为T。
对于顶点属于S时所产生的费用，只要从每个模块向t连一条容量为A[i]的边就可以对应起来，对于属于T的，只要从s向每个模块连一条容量为B[i]的边即可，接下对那些a[i]属于S，b[i]属于T时所产生的费用，只要从模块a[i]向模块b[i]连一条容量为w[i]的边就可以对应，对剩余情况同理可得。
code：
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAX=200010;
const int INF=1<<30;
struct edge{int to,cap,rev;};
vector<edge> G[MAX];
int level[MAX];
int iter[MAX];
void add(int from,int to,int cap)
{
G[from].push_back((edge){to,cap,G[to].size()});
G[to].push_back((edge){from,0,G[from].size()-1});
}
void bfs(int s)
{
memset(level,-1,sizeof(level));
queue<int> que;
level[s]=0;
que.push(s);
while(!que.empty())
{
int v=que.front();
que.pop();
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&& level[e.to]<0)
{
level[e.to]=level[v]+1;
que.push(e.to);
}
}
}
}
int dfs(int v,int t,int f)
{
if(v==t) return f;
for(int &i=iter[v];i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&&level[v]<level[e.to])
{
int d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
G[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int flow=0;
for(;;)
{
bfs(s);
if(level[t]<0) return flow;
memset(iter,0,sizeof(iter));
int f;
while((f=dfs(s,t,INF))>0)
flow+=f;
}
}
const int MAXV=20010;
int N,M;
int A[MAXV],B[MAXV];
int a[MAX],b[MAX],w[MAX];
void solve()
{
int s=N,t=s+1;
for(int i=0;i<N;i++)
{
add(i,t,A[i]);
add(s,i,B[i]);
}
for(int i=0;i<M;i++)
{
add(a[i]-1,b[i]-1,w[i]);
add(b[i]-1,a[i]-1,w[i]);
}
cout<<max_flow(s,t)<<endl;
}
int main()
{
cin>>N>>M;
for(int i=0;i<N;i++)
{
scanf("%d%d",&A[i],&B[i]);
}
for(int i=0;i<M;i++)
{
scanf("%d%d%d",&a[i],&b[i],&w[i]);
}
solve();
return 0;
}
poj3469(最大流最小割问题)