kuangbin专题专题十一 网络流 POJ 3436 ACM Computer Factory

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <queue>
  5 #include <vector>
  6 using namespace std;
  7 
  8 const int N = 110,INF = (int)1e9;
  9 int p,n,tot;
 10 int G[N][N],head[N],lev[N];
 11 queue<int > que;
 12 struct info{
 13     int in[20],out[20];
 14     int cap;
 15 }info[N];//存机器的信息
 16 struct node{
 17     int to,nxt,flow;
 18 }e[N*N];
 19 
 20 void init_main(){
 21     for(int i = 1; i <= n; ++i)
 22         for(int j = 1; j <= n; ++j) G[i][j] = 0;
 23 
 24     for(int i = 0; i <= 2*n+1; ++i) head[i] = -1; tot = 0;
 25 }
 26 
 27 void init_bfs(){
 28     for(int i = 1; i <= 2*n+1; ++i) lev[i] = 0;
 29     while(!que.empty()) que.pop();
 30 }
 31 
 32 inline void add(int u,int v,int flow){
 33     e[tot].to = v;
 34     e[tot].flow = flow;
 35     e[tot].nxt = head[u];
 36     head[u] = tot++;
 37 }
 38 
 39 //是否可连边
 40 inline bool check(int x,int y){
 41     for(int i = 1; i <= p; ++i){
 42         if(info[y].in[i] == 2) continue;
 43         if(info[x].out[i] != info[y].in[i]) return false;
 44     }
 45     return true;
 46 }
 47 
 48 //建边
 49 void rebuild(){
 50     for(int i = 1; i <= n; ++i){
 51         if(check(0,i)){
 52             add(0,i,INF); add(i,0,0);
 53         }
 54         if(check(i,2*n+1)){
 55             add(i+n,2*n+1,INF); add(2*n+1,i+n,0);
 56         }
 57     }
 58     for(int i = 1; i <= n; ++i){
 59         for(int j = 1; j <= n; ++j){
 60             if(i == j){
 61                 add(i,i+n,info[i].cap); add(i+n,i,0);
 62             }
 63             else if(check(i,j)){
 64                 add(i+n,j,INF); add(j,i+n,0);
 65             }
 66         }
 67     }
 68 }
 69 
 70 int dfs(int now,int flow,int t){
 71     if(now == t) return flow;
 72     int to,sum = 0,tmp = 0;
 73     for(int o = head[now]; ~o; o = e[o].nxt){
 74         to = e[o].to;
 75         if((lev[to] == lev[now] +1) && e[o].flow && (tmp = dfs(to,min(flow - sum, e[o].flow),t))){
 76             //需要的路径流量  G数组来存机器之间的联系
 77             if(now > n && now < 2*n+1 && to != 2*n+1){
 78                 G[now-n][to] += tmp;
 79             }
 80             e[o].flow -= tmp;
 81             e[o^1].flow += tmp;
 82             if((sum += tmp) == flow) return sum;
 83         }
 84     }
 85     return sum;
 86 }
 87 
 88 bool bfs(int s,int t){
 89     init_bfs();
 90     que.push(0);
 91     while(!que.empty()){
 92         int now = que.front(); que.pop();
 93         for(int o = head[now]; ~o; o = e[o].nxt){
 94             int to = e[o].to;
 95             if(!lev[to] && e[o].flow){
 96                 lev[to] = lev[now] + 1;
 97                 que.push(to);
 98             }
 99         }
100     }
101     if(lev[t]) return true;
102     else return false;
103 }
104 
105 int mf(int s,int t){
106     int max_flow = 0;
107     while(bfs(s,t)){
108         max_flow += dfs(s,INF,t);
109         //cout << "max_flow  " << max_flow << endl;
110     }
111     return max_flow;
112 }
113 
114 int main(){
115 
116     while(~scanf("%d%d",&p,&n)){
117         init_main();
118         //读入信息  0超级源点   2*n+1超级汇点
119         for(int i = 1; i <= n; ++i){
120             scanf("%d",&info[i].cap);
121             for(int j = 1; j <= p; ++j) scanf("%d",&info[i].in[j]);
122             for(int j = 1; j <= p; ++j) scanf("%d",&info[i].out[j]);
123         }
124         info[0].cap = INF; info[2*n+1].cap = INF;
125         for(int i = 1; i <= p; ++i){
126             info[0].out[i] = 0;
127             info[2*n+1].in[i] = 1;
128         }
129         //建图
130         rebuild();
131         //Dinic
132         int _mf = mf(0,2*n+1),line = 0;
133         //统计需要的联系数量
134         for(int i = 1; i <= n; ++i){
135             for(int j = 1; j <= n; ++j){
136                 if(G[i][j]) ++line;
137             }
138         }
139         printf("%d %d\n",_mf,line);
140         //输出联系
141         for(int i = 1; i <= n; ++i){
142             for(int j = 1; j <= n; ++j){
143                 if(G[i][j]) printf("%d %d %d\n",i,j,G[i][j]);
144             }
145         }
146      }
147 
148     return 0;
149 }