ACM Computer Factory
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 4077 | Accepted: 1366 | Special Judge | ||
Description
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 P numbers 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,P Di,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 1
Sample 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 0
Hint
Bold texts appearing in the sample sections are informative and do not form part of the actual data.
Source
Northeastern Europe 2005, Far-Eastern Subregion 做网络流的题目其实挺有意思的。重要的是建模。之后套模板了~~~~ 此题是最大流问题。题目意思比较难懂。 看得比较纠结。 就是说有N台组装电脑的机器。电脑的组成部分共有P部分。 每台机器有P个输入输出规格。还有一个容量Q；
其中输入规格有三种情况：0,1,2
0：该部分不能存在
1：该部分必须保留
2：该部分可有可无
输出规格有2种情况：0,1
0：该部分不存在
1：该部分存在
要求的是生产电脑最大的台数，就是求网络中的最大流。
这相当于是生产线。需要自己去建图。
但是考虑到每台机器都有容量，所以把一台机器分成两个点，中间建一条容量的边。
同时如果一台机器的输出符合另一台机器的输入，则建一条容量无穷大的边。
同时要增加源点和汇点。输入没有1的连接源点，输出全部是1的连接汇点。
程序：
/* POJ 3436 */ #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> #include<queue> using namespace std; //**************************************************** //最大流模板 //初始化：g[][],start,end //****************************************************** const int MAXN=110; const int INF=0x3fffffff; int g[MAXN][MAXN];//存边的容量，没有边的初始化为0 int path[MAXN],flow[MAXN],start,end; int n;//点的个数，编号0-n.n包括了源点和汇点。 queue<int>q; int bfs() { int i,t; while(!q.empty())q.pop();//把清空队列 memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1 path[start]=0; flow[start]=INF;//源点可以有无穷的流流进 q.push(start); while(!q.empty()) { t=q.front(); q.pop(); if(t==end)break; //枚举所有的点，如果点的编号起始点有变化可以改这里 for(i=0;i<=n;i++) { if(i!=start&&path[i]==-1&&g[t][i]) { flow[i]=flow[t]<g[t][i]?flow[t]:g[t][i]; q.push(i); path[i]=t; } } } if(path[end]==-1)return -1;//即找不到汇点上去了。找不到增广路径了 return flow[end]; } int Edmonds_Karp() { int max_flow=0; int step,now,pre; while((step=bfs())!=-1) { max_flow+=step; now=end; while(now!=start) { pre=path[now]; g[pre][now]-=step; g[now][pre]+=step; now=pre; } } return max_flow; } int backup[MAXN][MAXN];//备份图 int in[MAXN][20];//输入信息 int Line[MAXN][4]; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int p; int N; while(scanf("%d%d",&p,&N)!=EOF) { memset(g,0,sizeof(g)); for(int i=1;i<=N;i++) { for(int j=0;j<2*p+1;j++) scanf("%d",&in[i][j]); } for(int i=1;i<=N;i++)//拆点 { g[2*i-1][2*i]=in[i][0]; } n=2*N+1; start=0;//源点 end=n;//汇点 for(int i=1;i<=N;i++) { bool flag_s=true; bool flag_t=true; for(int j=0;j<p;j++) { if(in[i][j+1]==1)flag_s=false;//不能有1 if(in[i][j+1+p]==0)flag_t=false; } if(flag_s)g[0][2*i-1]=INF; if(flag_t)g[2*i][n]=INF; for(int j=1;j<=N;j++) if(i!=j) { bool flag=true; for(int k=0;k<p;k++) if((in[i][k+p+1]==0&&in[j][k+1]==1)||(in[i][k+p+1]==1&&in[j][k+1]==0)) { flag=false; break; } if(flag)g[2*i][2*j-1]=min(in[i][0],in[j][0]); } } memcpy(backup,g,sizeof(g));//先把图备份下来 printf("%d ",Edmonds_Karp()); int tol=0; for(int i=1;i<=N;i++) for(int j=1;j<=N;j++) { if(g[2*i][2*j-1]<backup[2*i][2*j-1]) { Line[tol][0]=i; Line[tol][1]=j; Line[tol][2]=backup[2*i][2*j-1]-g[2*i][2*j-1]; tol++; } } printf("%d\n",tol); for(int i=0;i<tol;i++) { printf("%d %d %d\n",Line[i][0],Line[i][1],Line[i][2]); } } return 0; }
另外再附上一个可以AC的代码。
没有拆点，取了最大值。
但是我认为是错误的，因为数据弱，所以AC了
/* POJ 3436 */ #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> #include<queue> using namespace std; //**************************************************** //最大流模板 //初始化：g[][],start,end //****************************************************** const int MAXN=110; const int INF=0x3fffffff; int g[MAXN][MAXN];//存边的容量，没有边的初始化为0 int path[MAXN],flow[MAXN],start,end; int n;//点的个数，编号0-n.n包括了源点和汇点。 queue<int>q; int bfs() { int i,t; while(!q.empty())q.pop();//把清空队列 memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1 path[start]=0; flow[start]=INF;//源点可以有无穷的流流进 q.push(start); while(!q.empty()) { t=q.front(); q.pop(); if(t==end)break; //枚举所有的点，如果点的编号起始点有变化可以改这里 for(i=0;i<=n;i++) { if(i!=start&&path[i]==-1&&g[t][i]) { flow[i]=flow[t]<g[t][i]?flow[t]:g[t][i]; q.push(i); path[i]=t; } } } if(path[end]==-1)return -1;//即找不到汇点上去了。找不到增广路径了 return flow[end]; } int Edmonds_Karp() { int max_flow=0; int step,now,pre; while((step=bfs())!=-1) { max_flow+=step; now=end; while(now!=start) { pre=path[now]; g[pre][now]-=step; g[now][pre]+=step; now=pre; } } return max_flow; } int backup[MAXN][MAXN];//备份图 int in[MAXN][20];//输入信息 int Line[MAXN][4]; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int p; while(scanf("%d%d",&p,&n)!=EOF) { memset(g,0,sizeof(g)); for(int i=1;i<=n;i++) { for(int j=0;j<2*p+1;j++) scanf("%d",&in[i][j]); } n++; start=0;//源点 end=n;//汇点 for(int i=1;i<=n-1;i++) { bool flag_s=true; bool flag_t=true; for(int j=0;j<p;j++) { if(in[i][j+1]==1)flag_s=false;//不能有1 if(in[i][j+1+p]==0)flag_t=false; } if(flag_s)g[0][i]=in[i][0]; if(flag_t)g[i][n]=in[i][0]; for(int j=1;j<=n-1;j++) if(i!=j) { bool flag=true; for(int k=0;k<p;k++) if((in[i][k+p+1]==0&&in[j][k+1]==1)||(in[i][k+p+1]==1&&in[j][k+1]==0)) { flag=false; break; } if(flag)g[i][j]=min(in[i][0],in[j][0]); } } memcpy(backup,g,sizeof(g));//先把图备份下来 printf("%d ",Edmonds_Karp()); int tol=0; for(int i=1;i<=n-1;i++) for(int j=1;j<=n-1;j++) { if(g[i][j]<backup[i][j]) { Line[tol][0]=i; Line[tol][1]=j; Line[tol][2]=backup[i][j]-g[i][j]; tol++; } } printf("%d\n",tol); for(int i=0;i<tol;i++) { printf("%d %d %d\n",Line[i][0],Line[i][1],Line[i][2]); } } return 0; }
