| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 5949 | Accepted: 2053 | 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
注意：本题Sample I/O这段英文不用输入输出
Sample input:
P N （N台机器，每台机器有P部分）
接着输入N行，其实每行都是一个结点的信息
每一行的格式为 一个Q P个S P个D
其中Q为当前结点的容量，S都是当前结点的输入规格，D都是输出规格
Sample output:
第一行的两个数字分别表示：最大流的值，流量发生变化的边数M（和s还有t关联的边不在其内，那些不属于原有的边，是附加边）
接下来有M行，每一行都有三个数字，A B W
A B为流量发生变化的边的端点，W为流量的变化值（每条边初始流量为0，最终流量就是找到最大流时的流量）
若图不连通，则输出0 0
1 #include <iostream> 2 #include<stdio.h> 3 #include<queue> 4 #include<string.h> 5 using namespace std; 6 int map[53][53],mapbk[53][53]; 7 int input[52][25]; 8 int path[53]; 9 int flow[53]; 10 int change[53][3]; 11 int start,end; 12 int p,n; 13 queue<int> q; 14 int BFS(){ 15 memset(path,-1,sizeof(path)); 16 while(!q.empty()) q.pop(); 17 q.push(start); 18 flow[start]=1000000; 19 path[start]=0; 20 while(!q.empty()){ 21 int v=q.front(); 22 if(v==end) 23 break; 24 q.pop(); 25 for(int i=0;i<=end;i++){ 26 if(path[i]==-1 && map[v][i]!=0 && start!=i){ 27 flow[i]=flow[v]<map[v][i]?flow[v]:map[v][i]; 28 path[i]=v; 29 q.push(i); 30 } 31 } 32 33 } 34 if(path[end]==-1) return -1; 35 else 36 return flow[end]; 37 } 38 int Edmonds_Karp(){ 39 int step,max_flow=0,now,pre; 40 while(1){ 41 step=BFS(); 42 if(step==-1) 43 break; 44 max_flow+=step; 45 now=end; 46 while(now!=start){ 47 pre=path[now]; 48 map[pre][now]-=step; 49 map[now][pre]+=step; 50 now=pre; 51 } 52 53 } 54 return max_flow; 55 } 56 int main() { 57 while(cin>>p>>n){ 58 memset(map,0,sizeof(map)); 59 memset(input,0,sizeof(input)); 60 //gets(str); 61 for(int i=1;i<=n;i++){ 62 for(int j=0;j<2*p+1;j++){ 63 int t; 64 cin>>t; 65 input[i][j]=t; 66 } 67 } 68 //getchar(); 69 // gets(str); 70 for(int i=1;i<=n;i++){ 71 int flag=1; 72 for(int j=1;j<p+1;j++){ 73 if(input[i][j]==1) 74 flag=0; 75 } 76 if(flag!=0){ 77 map[0][i]=input[i][0]; 78 } 79 flag=1; 80 for(int j=p+1;j<2*p+1;j++){ 81 if(input[i][j]==0) 82 flag=0; 83 } 84 if(flag!=0) 85 map[i][n+1]=input[i][0]; 86 flag=1; 87 for(int j=1;j<=n;j++){ 88 if(i!=j){ 89 for(int k=1;k<p+1;k++){ 90 if(i!=j&&input[i][p+k]+input[j][k]==1) 91 flag=0; 92 } 93 if(flag!=0) 94 map[i][j]=input[i][0]<input[j][0]?input[i][0]:input[j][0]; 95 flag=1; 96 } 97 } 98 99 } 100 start=0; 101 end=n+1; 102 memcpy(mapbk,map,sizeof(map)); 103 int result=Edmonds_Karp(); 104 105 int count=0; 106 for(int i=1;i<n+1;i++){ 107 for(int j=1;j<n+1;j++){ 108 if(map[i][j]<mapbk[i][j]){ 109 change[count][0]=i; 110 change[count][1]=j; 111 change[count][2]=mapbk[i][j]-map[i][j]; 112 count++; 113 } 114 } 115 } 116 cout<<result<<' '<<count<<endl; 117 for(int i=0;i<count;i++){ 118 cout<<change[i][0]<<' '<<change[i][1]<<' '<<change[i][2]<<endl; 119 } 120 } 121 return 0; 122 }