POJ 3436 最大流

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;    // vertex
    int cap;    // capacity
    int flow;   // current flow in this arc
    int 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0
    edge[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;
}