很羞愧啊 ， 题都没看咋懂
翻译见：http://hi.baidu.com/lewutian/blog/item/1b0709220085b7fed7cae28b.html/cmtid/87a1a0ad1aebe6064b36d60e
题解参考了：http://zhyu.me/acm/poj-3436.html ，BUPT一位现役神牛的blog
每台机器分输入输出，很显然直接拆点，每个点的(in)和(out)间连一条容量为该机器产量的边，然后对于输入没有限制的点，从源点向该点的(in)连一条容量为inf的边，对于输出是P个零件都有的点，从该点的(out)向汇点连一条容量inf的边，然后枚举所有点，如果某点的输出满足另一点的输入，则连一条inf的边。建图后直接求最大流即可，要输出解集，只需要最后遍历所有点的(out) ，看其所连的所有边中不包括终点的边，且流量大于0的边即为所求。
/*
ID: CUGB-wwj
PROG:
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define INF 1111111111
#define MAXN 222
#define MAXM 444444
#define PI acos(-1.0)
using namespace std;
struct node
{int ver; // vertexint cap; // capacityint flow; // current flow in this arcint next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e, p;
int ans[MAXM][3];
struct abcd
{int val;int in[15],out[15];
}x[MAXN];
int srcout[15], desin[15];
void add(int x, int y, int c)
{ //e记录边的总数edge[e].ver = y;edge[e].cap = c;edge[e].flow = 0;edge[e].rev = e + 1; //反向边在edge中的下标位置edge[e].next = head[x]; //记录以x为起点的上一条边在edge中的下标位置head[x] = e++; //以x为起点的边的位置//反向边edge[e].ver = x;edge[e].cap = 0; //反向边的初始网络流为0edge[e].flow = 0;edge[e].rev = e - 1;edge[e].next = head[y];head[y] = e++;
}
void rev_BFS()
{int Q[MAXN], qhead = 0, qtail = 0;for(int i = 1; i <= n; ++i){dist[i] = MAXN;numbs[i] = 0;}Q[qtail++] = des;dist[des] = 0;numbs[0] = 1;while(qhead != qtail){int v = Q[qhead++];for(int i = head[v]; i != -1; i = edge[i].next){if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;dist[edge[i].ver] = dist[v] + 1;++numbs[dist[edge[i].ver]];Q[qtail++] = edge[i].ver;}}
}
void init()
{e = 0;memset(head, -1, sizeof(head));
}
int maxflow()
{int u, totalflow = 0;int Curhead[MAXN], revpath[MAXN];for(int i = 1; i <= n; ++i)Curhead[i] = head[i];u = src;while(dist[src] < n){if(u == des) // find an augmenting path{int augflow = INF;for(int i = src; i != des; i = edge[Curhead[i]].ver)augflow = min(augflow, edge[Curhead[i]].cap);for(int i = src; i != des; i = edge[Curhead[i]].ver){edge[Curhead[i]].cap -= augflow;edge[edge[Curhead[i]].rev].cap += augflow;edge[Curhead[i]].flow += augflow;edge[edge[Curhead[i]].rev].flow -= augflow;}totalflow += augflow;u = src;}int i;for(i = Curhead[u]; i != -1; i = edge[i].next)if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;if(i != -1) // find an admissible arc, then Advance{Curhead[u] = i;revpath[edge[i].ver] = edge[i].rev;u = edge[i].ver;}else // no admissible arc, then relabel this vertex{if(0 == (--numbs[dist[u]]))break; // GAP cut, Important!Curhead[u] = head[u];int mindist = n;for(int j = head[u]; j != -1; j = edge[j].next)if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);dist[u] = mindist + 1;++numbs[dist[u]];if(u != src)u = edge[revpath[u]].ver; // Backtrack}}return totalflow;
}bool check(int* a, int* b)
{for(int i = 0; i < p; i++) if(a[i] + b[i] == 1) return 0;return 1;
}
int main()
{init();int nt;scanf("%d%d", &p, &nt);n = 2 * nt + 2;src = 1;des = n;for(int i = 0; i < p; i++){srcout[i] = 0;desin[i] = 1;}for(int i = 1; i <= nt; i++){scanf("%d", &x[i].val);for(int j = 0; j < p; j++) scanf("%d", &x[i].in[j]);for(int j = 0; j < p; j++) scanf("%d", &x[i].out[j]);}for(int i = 1; i <= nt; i++){if(check(srcout, x[i].in)) add(src, i + 1, INF);if(check(x[i].out, desin)) add(i + 1 + nt, des, INF);add(i + 1, i + nt + 1, x[i].val);for(int j = 1; j <= nt; j++) if(i != j && check(x[i].out, x[j].in)) add(i + 1 + nt, j + 1, INF);}rev_BFS();printf("%d ", maxflow());int cnt = 0;for(int i = nt + 2; i <= des - 1; i ++)for(int j = head[i]; j != -1; j = edge[j].next){if(edge[j].ver != des && edge[j].flow > 0){ans[cnt][0] = i - nt - 1;ans[cnt][1] = edge[j].ver - 1;ans[cnt++][2] = edge[j].flow;}}printf("%d\n", cnt);for(int i = 0; i < cnt; i++)printf("%d %d %d\n", ans[i][0], ans[i][1], ans[i][2]);return 0;
}