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-wwjPROG: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;}