POJ 3469 Dual Core CPU

最新推荐文章于 2018-07-17 10:38:00 发布

ShinyaLicone

最新推荐文章于 2018-07-17 10:38:00 发布

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分类专栏： POJ 网络流文章标签： C++ 最小割

本文链接：https://blog.youkuaiyun.com/SenyeLicone/article/details/54666829

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本文介绍了一种针对双核CPU环境下，通过最小割算法来优化任务分配的问题解决方法。该算法旨在降低运行成本，特别是在模块间需要进行数据交换的情况下。通过对任务与核心之间的连接建立模型，并考虑额外的数据交换成本，最终实现总成本最小化。

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 A_i and B_i. Meanwhile, M pairs of modules need to do some data-exchange. If they are running on the same core, then the cost of this action can be ignored. Otherwise, some extra cost are needed. You should arrange wisely to minimize the total cost.

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, A_i and B_i.
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 between them.

Output

Output only one integer, the minimum total cost.

Sample Input

Sample Output

Source

POJ Monthly--2007.11.25, Zhou Dong

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

最小割~

从S向所有任务连边表示A机器，从所有任务向T连边表示B机器，然后对于每一对x,y连双向边，表示如果选择不同就要花费权值，跑最小割即可~

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;

int n,m,x,y,val,fi[20010],w[4001000],ne[4001000],v[4001000],dis[20010],cnt,ans;

void add(int u,int vv,int val)
{
	w[++cnt]=vv;ne[cnt]=fi[u];fi[u]=cnt;v[cnt]=val;
	w[++cnt]=u;ne[cnt]=fi[vv];fi[vv]=cnt;v[cnt]=0;
}

bool bfs()
{
	queue<int> q;
	memset(dis,-1,sizeof(dis));
	q.push(0);dis[0]=0;
	while(!q.empty())
	{
		int k=q.front();q.pop();
		for(int i=fi[k];i;i=ne[i])
		  if(v[i]>0 && dis[w[i]]==-1)
		  {
		  	dis[w[i]]=dis[k]+1;q.push(w[i]);
		  }
	}
	if(dis[n+1]==-1) return 0;
	return 1;
}

int findd(int u,int vv)
{
	if(u==n+1) return vv;
	int kkz,tot=0;
	for(int i=fi[u];i;i=ne[i])
	  if(v[i]>0 && dis[w[i]]==dis[u]+1 && (kkz=findd(w[i],min(v[i],vv-tot))))
	  {
	  	v[i]-=kkz;v[i^1]+=kkz;tot+=kkz;
	  	if(tot==vv) return tot;
	  }
	if(!tot) dis[u]=-1;
	return tot;
}

int main()
{
	scanf("%d%d",&n,&m);cnt=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&x,&y);
		add(0,i,x);add(i,n+1,y);
	}
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&val);
		add(x,y,val);add(y,x,val);
	}
	while(bfs()) ans+=findd(0,999999999);
	printf("%d\n",ans);
	return 0;
}