POJ 3469 Dual Core CPU
来源：程序员人生 发布时间：2014-10-02 08:00:01 阅读次数：2424次
一堆任务分配到2个不同的芯片上运行,一个任务在不同芯片上运行的时间不一样,有一些任务组如果分配到不同的芯片上运行会产生额外的时间....
用最小的费用将不同对象划分到两个集合 , 最小割问题 .
建图:
用源点汇点代表两个芯片
对某个任务 , 在 A 芯片上运行时间 t1 . 则从源点连一条 到 这个任务容量为 t1 的边 . 对 B 芯片上运行时间同理
如果两个任务间有联系,着再俩个任务间连边.
原问题就转化成了, 将图分成两部分,所切割的容量最少.
最小割==最大流 isap 解决
|
Dual Core CPU
| Time Limit: 15000MS |
|
Memory Limit: 131072K |
| Total Submissions: 19127 |
|
Accepted: 8263 |
| Case Time Limit: 5000MS |
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
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, 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
between them.
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
|
[Submit] [Go Back]
[Status] [Discuss]
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=20200;
const int maxm=600300;
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow;
}edge[maxm];
int Size,Adj[maxn];
int gap[maxn],dep[maxn],pre[maxn],cur[maxn];
void init()
{
Size=0;
memset(Adj,-1,sizeof(Adj));
}
void addedge(int u,int v,int w,int rw=0)
{
edge[Size].to=v; edge[Size].cap=w; edge[Size].next=Adj[u];
edge[Size].flow=0; Adj[u]=Size++;
edge[Size].to=u; edge[Size].cap=rw; edge[Size].next=Adj[v];
edge[Size].flow=0; Adj[v]=Size++;
}
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,Adj,sizeof(Adj));
int u=start;
pre[u]=-1;gap[0]=N;
int ans=0;
while(dep[start]<N)
{
if(u==end)
{
int Min=INF;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
if(Min>edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];~i;i=edge[i].next)
{
v=edge[i].to;
if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag)
{
u=v;
continue;
}
int Min=N;
for(int i=Adj[u];~i;i=edge[i].next)
{
if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if(u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
int n,m;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
for(int i=1;i<=n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
addedge(0,i,a);
addedge(i,n+1,b);
}
for(int i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w,w);
}
printf("%d
",sap(0,n+1,n+2));
}
return 0;
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
