As you know, all the computers used for ACM contests must be identical, so the participants compete on equal terms. That is why all these computers are historically produced at the same factory.
Every ACM computer consists of P parts. When all these parts are present, the computer is ready and can be shipped to one of the numerous ACM contests.
Computer manufacturing is fully automated by using N various machines. Each machine removes some parts from a half-finished computer and adds some new parts (removing of parts is sometimes necessary as the parts cannot be added to a computer in arbitrary order). Each machine is described by its performance (measured in computers per hour), input and output specification.
Input specification describes which parts must be present in a half-finished computer for the machine to be able to operate on it. The specification is a set of P numbers 0, 1 or 2 (one number for each part), where 0 means that corresponding part must not be present, 1 — the part is required, 2 — presence of the part doesn't matter.
Output specification describes the result of the operation, and is a set of Pnumbers 0 or 1, where 0 means that the part is absent, 1 — the part is present.
The machines are connected by very fast production lines so that delivery time is negligibly small compared to production time.
After many years of operation the overall performance of the ACM Computer Factory became insufficient for satisfying the growing contest needs. That is why ACM directorate decided to upgrade the factory.
As different machines were installed in different time periods, they were often not optimally connected to the existing factory machines. It was noted that the easiest way to upgrade the factory is to rearrange production lines. ACM directorate decided to entrust you with solving this problem.
Input
Input file contains integers P N, then N descriptions of the machines. The description of ith machine is represented as by 2 P + 1 integers Qi Si,1 Si,2...Si,PDi,1 Di,2...Di,P, where Qi specifies performance, Si,j — input specification for part j, Di,k — output specification for part k.
Constraints
1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Qi ≤ 10000
Output
Output the maximum possible overall performance, then M — number of connections that must be made, then M descriptions of the connections. Each connection between machines A and B must be described by three positive numbers A B W, where W is the number of computers delivered from A to B per hour.
If several solutions exist, output any of them.
Sample Input
Sample input 1
3 4
15 0 0 0 0 1 0
10 0 0 0 0 1 1
30 0 1 2 1 1 1
3 0 2 1 1 1 1
Sample input 2
3 5
5 0 0 0 0 1 0
100 0 1 0 1 0 1
3 0 1 0 1 1 0
1 1 0 1 1 1 0
300 1 1 2 1 1 1
Sample input 3
2 2
100 0 0 1 0
200 0 1 1 1Sample Output
Sample output 1
25 2
1 3 15
2 3 10
Sample output 2
4 5
1 3 3
3 5 3
1 2 1
2 4 1
4 5 1
Sample output 3
0 0Hint
Bold texts appearing in the sample sections are informative and do not form part of the actual data.
很裸的一道题
网络流首发ac
其实dinic 代码并不算长
本体难在建模 只要有了想法 把图建好
那么我们直接用网络流模板dinic就可以了
我们只需要将每个机器拆成两割点 在中间连一条边
这条边 的流量代表效率
然后再暴力匹配 把每一个能匹配的机器建立单向边
那么就可以很轻易的ac了
#include <iostream>
#include <queue>
#include<map>
#include<cstring>
#include<cstdio>
using namespace std;
const int INF = 0x7fffffff;
int p,n;
int level[205];
int Si, Ei, Ci;
int node[105][55];
int va[105];
int vis[105][105];
int to[105][3];
struct Dinic
{
int c;
int f;
} edge[205][205];
bool dinic_bfs() //bfs方法构造层次网络
{
//cout<<"level"<<endl;
queue<int> q;
memset(level, 0, sizeof(level));
q.push(0);
level[0] = 1;
int u, v;
while (!q.empty())
{
u = q.front();
q.pop();
for (v = 1; v <= 2*n+1; v++)
{
if (!level[v] && edge[u][v].c>edge[u][v].f)
{
level[v] = level[u] + 1;
q.push(v);
}
}
}
return level[2*n+1] != 0; //question: so it must let the sink node is the Mth?/the way of yj is give the sink node's id
}
int dinic_dfs(int u, int cp) //use dfs to augment the flow
{
int tmp = cp;
int v, t;
if (u == 2*n+1)
return cp;
for (v = 1; v <= 2*n+1&&tmp; v++)
{
if (level[u] + 1 == level[v])
{
if (edge[u][v].c>edge[u][v].f)
{
t = dinic_dfs(v, min(tmp, edge[u][v].c - edge[u][v].f));
edge[u][v].f += t;
edge[v][u].f -= t;
tmp -= t;
}
}
}
return cp - tmp;
}
int dinic()
{
int sum=0, tf=0;
while (dinic_bfs())
{
while (tf = dinic_dfs(0, INF))
sum += tf;
}
return sum;
}
int main()//0为起点 2n+1为终点 奇数为边的起点 偶数为边的终点
{
scanf("%d%d",&p,&n);
memset(vis,0,sizeof(vis));
memset(node,0,sizeof(node));
memset(edge,0,sizeof(edge));
for(int i=1;i<=p;i++) node[2*n+1][i]=1;
for(int i=1;i<=n;i++)
{
scanf("%d",&edge[2*i-1][2*i].c);
for(int j=1;j<=p;j++)
scanf("%d",&node[2*i-1][j]);
for(int j=1;j<=p;j++)
scanf("%d",&node[2*i][j]);
}
for(int i=1;i<=n;i++)//起点为奇数
{
for(int j=0;j<=n;j++)//重点为偶数
{
int tmp1=i*2-1;
int tmp2=j*2;
int flag=0;
for(int k=1;k<=p;k++)
{
if(node[tmp1][k]!=2&&node[tmp1][k]!=node[tmp2][k])
{
flag=1;
}
}
if(flag==0)
{
edge[tmp2][tmp1].c=1e9;
vis[tmp2][tmp1]=1;
}
}
}
for(int i=1;i<=n;i++)
{
int flag=0;
int tmp1=i*2;
for(int j=1;j<=p;j++)
{
if(node[tmp1][j]!=1)
{
flag=1;
break;
}
}
if(flag==0) edge[tmp1][2*n+1].c=1e9;
}
// for(int i=0;i<=2*n+1;i++)
// {
// for(int j=0;j<=2*n+1;j++)
// {
// printf("%10d ",edge[i][j].c);
// }
// printf("
");
// }
int ans=dinic();
int num=0;
for(int i=1;i<=n;i++)//由一个终点到达另外的一个起点
{
for(int j=1;j<=n+1;j++)
{
int tmp1=2*i;
int tmp2=2*j-1;
if(vis[tmp1][tmp2]==1&&edge[tmp1][tmp2].f>0)
{
num++;
to[num][0]=i;
to[num][1]=j;
to[num][2]=edge[tmp1][tmp2].f;
}
}
}
printf("%d %d
",ans,num);
for(int i=1;i<=num;i++)
{
printf("%d %d %d
",to[i][0],to[i][1],to[i][2]);
}
return 0;
}