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 asAi 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 extraw 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
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int inf=(1<<28);
using namespace std;
#define maxn 500000
struct edge
{
int u,v,next,f,pre;
}e[2*maxn];
int num,rnum;
int head[maxn],rhead[maxn];
int d[maxn];
int numb[maxn];
int start[maxn];
int n,m;//见图后的点数(1->n)和原图边数
int p[maxn];
int source,sink;
//要初始化source 和 sink,重定义n
void Init()
{
memset(head,-1,sizeof(head));
memset(rhead,-1,sizeof(rhead));
memset(p,-1,sizeof(p));
num=0;
return ;
}
void BFS()
{
int i,j;
for(i=1;i<=n;i++)
{
d[i]=n;
numb[i]=0;
}
int Q[maxn],head(0),tail(0);
d[sink]=0;
numb[0]=1;
Q[++tail]=sink;
while(head<tail)
{
i=Q[++head];
for(j=rhead[i];j!=-1;j=e[j].pre)
{
if(e[j].f==0||d[e[j].u]<n)
continue;
d[e[j].u]=d[i]+1;
numb[d[e[j].u]]++;
Q[++tail]=e[j].u;
}
}
return ;
}
int Augment()
{
int i;
int tmp=inf;
for(i=p[sink];i!=-1;i=p[e[i].u])
{
if(tmp>e[i].f)
tmp=e[i].f;
}
for(i=p[sink];i!=-1;i=p[e[i].u])
{
e[i].f-=tmp;
e[i^1].f+=tmp;
}
return tmp;
}
int Retreat(int &i)
{
int tmp,j,mind(n-1);
for(j=head[i];j!=-1;j=e[j].next)
{
if(e[j].f>0&&d[e[j].v]<mind)
mind=d[e[j].v];
}
tmp=d[i];
d[i]=mind+1;
numb[tmp]--;
numb[d[i]]++;
if(i!=source)
i=e[p[i]].u;
return numb[tmp];
}
int maxflow()
{
int flow(0),i,j;
BFS();
for(i=1;i<=n;i++)
start[i]=head[i];
i=source;
while(d[source]<n)
{
for(j=start[i];j!=-1;j=e[j].next)
if(e[j].f>0&&d[i]==d[e[j].v]+1)
break;
if(j!=-1)
{
start[i]=j;
p[e[j].v]=j;
i=e[j].v;
if(i==sink)
{
flow+=Augment();
i=source;
}
}
else
{
start[i]=head[i];
if(Retreat(i)==0)
break;
}
}
return flow;
}
//a->b=c;
void addedge(int a,int b,int c)
{
e[num].next=head[a];
head[a]=num;
e[num].pre=rhead[b];
rhead[b]=num;
e[num].f=c;
e[num].u=a;
e[num++].v=b;
e[num].next=head[b];
head[b]=num;
e[num].pre=rhead[a];
rhead[a]=num;
e[num].u=b;
e[num].v=a;
e[num++].f=0;
return ;
}
int main()
{
while(scanf("%d%d",&n,&m)==2)
{
n+=2;
source=1,sink=n;
Init();
for(int i=2;i<n;i++)
{
int a,b;scanf("%d%d",&a,&b);
addedge(source,i,a);
addedge(i,sink,b);
}
for(int i=1;i<=m;i++)
{
int u,v,w;scanf("%d%d%d",&u,&v,&w);u++,v++;
addedge(u,v,w);//此题中要加双向边
addedge(v,u,w);
}
int cnt=maxflow();
printf("%d\n",cnt);
}
return 0;
}