18.11.23 POJ 3436 ACM Computer Factory（dinic）

Posted 2021-01-22 yalphait

描述

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 Nvarious 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 Pnumbers 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.

输入

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 Q_i S_i,1 Si_,2...S_i,P D_i,1 D_i,2...D_i,P, where Q_ispecifies performance, S_i,j — input specification for part j, D_i,k — output specification for part k.

Constraints

1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Q_i ≤ 10000

输出

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 Aand 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.

样例输入

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

样例输出

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

提示

Bold texts appearing in the sample sections are informative and do not form part of the actual data.来源Northeastern Europe 2005, Far-Eastern Subregion

  1 #include <iostream>
  2 #include <string.h>
  3 #include <algorithm>
  4 #include <stack>
  5 #include <string>
  6 #include <math.h>
  7 #include <queue>
  8 #include <stdio.h>
  9 #include <string.h>
 10 #include <vector>
 11 #include <fstream>
 12 #include <set>
 13 #define maxn 105
 14 #define inf 0x7fffffff
 15 
 16 using namespace std;
 17 struct node {
 18     int oldstatus[11], newstatus[11];
 19     node() {
 20         memset(oldstatus, 0, sizeof(oldstatus));
 21         memset(newstatus, 0, sizeof(newstatus));
 22     }
 23 };
 24 node _node[maxn];
 25 int G[maxn][maxn],Q[maxn];
 26 int p, n;
 27 int Layer[maxn];
 28 int flow[maxn][maxn];
 29 
 30 bool countlayer() {
 31     int layer = 0;
 32     deque<int>q;
 33     memset(Layer, 0xff, sizeof(Layer));
 34     Layer[0] = 0;
 35     q.push_back(0);
 36     while (!q.empty()) {
 37         int v = q.front();
 38         q.pop_front();
 39         for (int j = 0; j <= n + 1; j++) {
 40             if (G[v][j] > 0 && Layer[j] == -1) {
 41                 Layer[j] = Layer[v] + 1;
 42                 if (j == n+1)return true;
 43                 else q.push_back(j);
 44             }
 45         }
 46     }
 47     return false;
 48 }
 49 
 50 bool visited[maxn];
 51 void dicnic() {
 52     int MaxFlow = 0;
 53     deque<int>q;
 54     while (countlayer()) {
 55         q.push_back(0);
 56         memset(visited, 0, sizeof(visited));
 57         visited[0] = true;
 58         while (!q.empty()) {
 59             int nd = q.back();
 60             if (nd == n + 1) {
 61                 int nMinC = inf, Min_vs;
 62                 for (int i = 1; i < q.size(); i++) {
 63                     int vs = q[i - 1], ve = q[i];
 64                     if (G[vs][ve] > 0) {
 65                         if (nMinC > G[vs][ve]) {
 66                             nMinC = G[vs][ve];
 67                             Min_vs = vs;
 68                         }
 69                     }
 70                 }
 71                 MaxFlow += nMinC;
 72                 for (int i = 1; i < q.size(); i++) {
 73                     int vs = q[i - 1], ve = q[i];
 74                     G[vs][ve] -= nMinC;
 75                     G[ve][vs] += nMinC;
 76                     flow[vs][ve] += nMinC;
 77                 }
 78                 while (!q.empty() && q.back() != Min_vs) {
 79                     visited[q.back()] = false;
 80                     q.pop_back();
 81                 }
 82             }
 83             else {
 84                 int i;
 85                 for (i = 0; i <= n + 1; i++) {
 86                     if (G[nd][i] > 0 && Layer[i] == Layer[nd] + 1 && !visited[i]) {
 87                         visited[i] = true;
 88                         q.push_back(i);
 89                         break;
 90                     }
 91                 }
 92                 if (i > n + 1)
 93                     q.pop_back();
 94             }
 95         }
 96     }
 97     printf("%d", MaxFlow);
 98 }
 99 
100 void init() {
101     scanf("%d%d", &p, &n);
102     for (int i = 1; i <= n; i++) {
103         scanf("%d", &Q[i]);
104         bool flag = true;
105         for (int j = 1; j <= p; j++)
106         {
107             scanf("%d", &_node[i].oldstatus[j]);
108             if (_node[i].oldstatus[j] == 1)
109                 flag = false;
110         }
111         if (flag)G[0][i] = Q[i];
112         flag = true;
113         for (int j = 1; j <= p; j++)
114         {
115             scanf("%d", &_node[i].newstatus[j]);
116             if (_node[i].newstatus[j] != 1)flag = false;
117         }
118         if (flag)G[i][n+1] = Q[i];
119     }
120     for(int i=1;i<=n;i++)
121         for (int j = 1; j <= n; j++) {
122             {
123                 if (i == j)continue;
124                 bool flag = true;
125                 for (int k = 1; k <= p; k++)
126                     if (_node[i].oldstatus[k] == 1 && _node[j].newstatus[k] == 0
127                         || _node[i].oldstatus[k] == 0 && _node[j].newstatus[k] == 1) {
128                         flag = false;
129                         break;
130                     }
131                 if (flag)
132                     G[j][i] = Q[j];
133             }
134         }
135     dicnic();
136     int sum = 0;
137     for(int i=1;i<=n;i++)
138         for (int j = 1; j <= n; j++) {
139             if (i == j)continue;
140             int res = flow[i][j]-flow[j][i];
141             if (res > 0)
142                 sum++;
143         }
144     printf(" %d
", sum);
145     for (int i = 1; i <= n; i++)
146         for (int j = 1; j <= n; j++) {
147             if (i == j)continue;
148             int res = flow[i][j] - flow[j][i];
149             if (res > 0)
150                 printf("%d %d %d
", i, j, res);
151         }
152 }
153 
154 int main()
155 {
156     init();
157     return 0;
158 }

View Code

其实是数据太水了，本来要拆点的，被我蒙过去了，其实不拆不行的（x

并且 我dinic写错了（x

写的时候懒了那么一下（并且被最后那张ppt迷惑了）