POJ 3436 ACM Computer Factory (最大流)

標籤：

ACM Computer Factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6773		Accepted: 2379		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 13 415  0 0 0  0 1 010  0 0 0  0 1 130  0 1 2  1 1 13   0 2 1  1 1 1Sample input 23 55   0 0 0  0 1 0100 0 1 0  1 0 13   0 1 0  1 1 01   1 0 1  1 1 0300 1 1 2  1 1 1Sample input 32 2100  0 0  1 0200  0 1  1 1

Sample Output

Sample output 125 21 3 152 3 10Sample output 24 51 3 33 5 31 2 12 4 14 5 1Sample output 30 0

分析
題意就是一個電腦有P個組件，有N台機器，每個機器只能加工指定組件存在的半成品電腦，然後會加上或者減去電腦的一些組件，每台機器有一個performance，最後就是求最大的performance的和

  其中輸入規格有三種情況：0,1,2

  0：該部分不能存在

  1：該部分必須保留

  2：該部分可有可無

  輸出規格有2種情況：0,1

  0：該部分不存在

  1：該部分存在

       就是求最大流，關鍵在於建圖，首先每個機器有一個performance，所以將每個機器拆成兩個點，中間用邊權為w的有向線段串連；

  輸入規格全是0的和源點相連，全是1的和匯點相連。兩台機器i和j，如果i的輸出符合j的輸入，就將他們連起來，取流量較小的一個

  然後跑最大流就行了。

  輸出路徑的話就先儲存沒跑最大流之前的圖，和之後的比較，如果小，就說明這兩個點是路徑上的點，存起來輸出就行了。

#include<stdio.h>#include<string.h>#include<iostream>#include<queue>using namespace std;//****************************************************//最大流模板Edmonds_Karp演算法//初始化：G[][],st,ed//******************************************************const int MAXN = 200+10;const int INF = 0x3fffffff;int G[MAXN][MAXN];//存邊的容量，沒有邊的初始化為0int path[MAXN],flow[MAXN],st,ed;int n;//點的個數，編號0~n，n包括了源點和匯點queue<int>q;int bfs(){    int i,t;    while(!q.empty()) q.pop();//清空隊列    memset(path,-1,sizeof(path));//每次搜尋前都把路徑初始化成-1    path[st]=0;    flow[st]=INF;//源點可以有無窮的流流進    q.push(st);    while(!q.empty()){        t=q.front();        q.pop();        if(t==ed) break;        for(i=0;i<=n;i++){            if(i!=st&&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[ed]==-1) return -1;//即找不到匯點上去了。找不到增廣路徑了    return flow[ed];}int Edmonds_Karp(){    int max_flow=0;    int step,now,pre;    while((step=bfs())!=-1){        max_flow+=step;        now=ed;        while(now!=st){            pre=path[now];            G[pre][now]-=step;            G[now][pre]+=step;            now=pre;        }    }    return max_flow;}int in[MAXN][20];//輸入資訊int backup[MAXN][MAXN];//備份圖int Line[MAXN][4];int main(){    int P,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++)//拆點            G[2*i-1][2*i]=in[i][0];        n=2*N+1;        st=0,ed=n;        for(int i=1;i<=N;i++){            bool flag_s=true;            bool flag_t=true;            for(int j=1;j<=P;j++){                if(in[i][j]==1) flag_s=false;                if(in[i][j+P]==0) flag_t=false;            }            if(flag_s) G[st][2*i-1]=INF;            if(flag_t) G[2*i][ed]=INF;            for(int j=1;j<=N;j++){                if(j==i) continue;                bool flag=true;                for(int k=1;k<=P;k++)                    if((in[i][k+P]==0&&in[j][k]==1)||(in[i][k+P]==1&&in[j][k]==0)){                       flag=false;                       break;                    }                if(flag) G[2*i][2*j-1]=min(in[i][0],in[j][0]);            }        }        memcpy(backup,G,sizeof(G));        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;}

 

 

POJ 3436 ACM Computer Factory (最大流)