POJ3436 ACM Computer Factory【网络流】

题意：有n台机器组装有p个零件的电脑，每台机器的输入参数为0、1、2，分别代表某个零件无、有、可有可无。输出参数为0、1，代表某个零件无、有。每台机器有加工上限，问最多能组装多少电脑

思路：最大流，n台机器拆点，入点、出点，容量为加工上限。当一台机器的输出全为1的，就是电脑组装完成，连汇点。当一台机器输入没有1的时候就是刚开始加工，与起点相连。机器和机器之间，输出和输入相匹配，连起来。除了上边说明的边的容量，其他全为inf

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<set>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;typedef pair<int,int> pii;const int maxn = 150;struct edge {int to,cf,rev;};vector<edge> G[maxn];int lev[maxn],iter[maxn];int in[maxn][maxn],out[maxn][maxn],answ[maxn*maxn][5];void init(int x){for(int i = 0; i <= x; i++)G[i].clear();}void add(int from, int to, int cap){G[from].push_back((edge){to,cap,G[to].size()});G[to].push_back((edge){from,0,G[from].size()-1});}void bfs(int s){memset(lev,-1,sizeof lev);queue<int> q;lev[s] = 0;q.push(s);while(!q.empty()){int v = q.front();q.pop();for(int i = 0; i < G[v].size(); i++){edge &e = G[v][i];if(e.cf > 0 && lev[e.to] < 0){lev[e.to] = lev[v] + 1;q.push(e.to);}}}}int dfs(int v,int t, int f){if(v == t) return f;for(int &i = iter[v]; i < G[v].size(); i++){edge &e = G[v][i];if(e.cf > 0 && lev[v] < lev[e.to]){int d = dfs(e.to, t, min(f,e.cf));if(d > 0){e.cf -= d;G[e.to][e.rev].cf += d;return d;}}}return 0;}int maxflow(int s,int t){int flow = 0;while(1){bfs(s);if(lev[t] < 0) return flow;memset(iter,0,sizeof iter);int f;while((f = dfs(s,t,0x3f3f3f3f)) > 0)flow += f;}}int check(int x,int y,int p){for(int i = 1; i <= p; i++){if(in[y][i] + out[x][i] == 1)return 0;}return 1;}int main(void){int p,n;while(scanf("%d%d",&p,&n)!=EOF){init(2*n+2);for(int i = 1; i <= n; i++){int k,num = 0;scanf("%d",&k);add(i,i+n,k);for(int j = 1; j <= p; j++){scanf("%d",&in[i][j]);if(in[i][j] == 1)num++;}if(!num) add(0,i,0x3f3f3f3f);num = 0;for(int j = 1; j <= p; j++){scanf("%d",&out[i][j]);if(out[i][j] == 0)num++;}if(!num) add(i+n,2*n+1,0x3f3f3f3f);}for(int i = 1; i <= n; i++){for(int j = i+1; j <= n; j++){if(check(i,j,p))add(i+n,j,0x3f3f3f3f);if(check(j,i,p))add(j+n,i,0x3f3f3f3f);}}int ans = maxflow(0,2*n+1),tot = 0;for(int i = 1+n; i <= 2*n; i++){for(int j = 0; j < G[i].size(); j++){if(G[i][j].cf < 0x3f3f3f3f && G[i][j].to != 2*n+1 && G[i][j].to != i-n){answ[tot][0] = i-n;answ[tot][1] = G[i][j].to;answ[tot++][2] = 0x3f3f3f3f - G[i][j].cf;}}}printf("%d %d\n",ans,tot);for(int i = 0; i < tot; i++)printf("%d %d %d\n",answ[i][0],answ[i][1],answ[i][2]);}return 0;}