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POJ 3436 ACM Computer Factory (拆点+输出解)

【题意】每台计算机由P个零件组成，工厂里有n台机器，每台机器针对P个零件有不同的输入输出规格，现在给出每台机器每小时的产量，问如何建立流水线（连接各机器）使得每小时生产的计算机最多。网络流的建图真的比较有意思~~ 【建图】如下图，建一个输出规格都为0的超级源点表示起始状态，建一个输入规格都为1的超级汇点表示完成一台电脑，把每一个机器拆成一个源点和一个汇点，之间连一条performance的边表示每小时能产几台。对于任意两台机器，如果其中一台的输出规格=另一台的输入规格，则两边连一条无穷流量的边（包括超级源汇点），求出最大流即可。

【输出解】这个比较好想，遍历流网络，可以只看反向边，如果反向边流量>0则表示该条边有流量，输出.

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int oo = 0x3fffffff;
const int MAXV = 205;
const int MAXE = 30005;
struct node{
    int u, v, flow;
    int opp;
    int next;
};
int inq[52][12];
int outq[52][12];
struct Dinic{
    node arc[MAXE];
    int vn, en, head[MAXV];     //vn点个数(包括源点汇点),en边个数
    int cur[MAXV];              //当前弧
    int q[MAXV];                //bfs建层次图时的队列
    int path[MAXE], top;        //存dfs当前最短路径的栈
    int dep[MAXV];              //各节点层次
    void init(int n){
        vn = n;
        en = 0;
        mem(head, -1);
    }
    void insert_flow(int u, int v, int flow){
        arc[en].u = u;
        arc[en].v = v;
        arc[en].flow = flow;
        arc[en].opp = en + 1;
        arc[en].next = head[u];
        head[u] = en ++;

        arc[en].u = v;
        arc[en].v = u;
        arc[en].flow = 0;       //反向弧
        arc[en].opp = en - 1;
        arc[en].next = head[v];
        head[v] = en ++;
    }
    bool bfs(int s, int t){
        mem(dep, -1);
        int lq = 0, rq = 1;
        dep[s] = 0;
        q[lq] = s;
        while(lq < rq){
            int u = q[lq ++];
            if (u == t){
                return true;
            }
            for (int i = head[u]; i != -1; i = arc[i].next){
                int v = arc[i].v;
                if (dep[v] == -1 && arc[i].flow > 0){
                    dep[v] = dep[u] + 1;
                    q[rq ++] = v;
                }
            }
        }
        return false;
    }
    int solve(int s, int t){
        int maxflow = 0;
        while(bfs(s, t)){
            int i, j;
            for (i = 1; i <= vn; i ++)  cur[i] = head[i];
            for (i = s, top = 0;;){
                if (i == t){
                    int mink;
                    int minflow = 0x3fffffff;
                    for (int k = 0; k < top; k ++)
                        if (minflow > arc[path[k]].flow){
                            minflow = arc[path[k]].flow;
                            mink = k;
                        }
                    for (int k = 0; k < top; k ++)
                        arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
                    maxflow += minflow;
                    top = mink;		//arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
                    i = arc[path[top]].u;
                }
                for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
                    int v = arc[j].v;
                    if (arc[j].flow && dep[v] == dep[i] + 1)
                        break;
                }
                if (j != -1){
                    path[top ++] = j;
                    i = arc[j].v;
                }
                else{
                    if (top == 0)   break;
                    dep[i] = -1;
                    i = arc[path[-- top]].u;
                }
            }
        }
        return maxflow;
    }
}dinic;
struct Path{
    int u, v, w;
}path[MAXE];
int main(){
    int n, p;
    while(scanf("%d %d", &p, &n) != EOF){
        dinic.init(2*(n+1));
        for (int i = 1; i <= n; i ++){
            int q;
            scanf("%d", &q);
            for (int j = 0; j < p; j ++)
                scanf("%d", &inq[i][j]);
            for (int j = 0; j < p; j ++)
                scanf("%d", &outq[i][j]);
            dinic.insert_flow(2*i-1, 2*i, q);
        }
        for (int i = 1; i <= n; i ++){
            int ok = 1;
            for (int j = 0; j < p; j ++){
                if (inq[i][j] == 1){
                    ok = 0;
                    break;
                }
            }
            if (ok){
                dinic.insert_flow(2*n+1, 2*i-1, oo);
            }
        }
         for (int i = 1; i <= n; i ++){
            int ok = 1;
            for (int j = 0; j < p; j ++){
                if (outq[i][j] == 0){
                    ok = 0;
                    break;
                }
            }
            if (ok){
                dinic.insert_flow(2*i, 2*n+2, oo);
            }
        }
        for (int i = 1; i <= n; i ++){
            for (int j = 1; j <= n; j ++){
                if (i != j){
                    int ok = 1;
                    for(int k = 0; k < p; k ++){
                        if (inq[j][k] == 2) continue;
                        if (outq[i][k] != inq[j][k]){
                            ok = 0;
                            break;
                        }
                    }
                    if (ok){
                        dinic.insert_flow(2*i, 2*j-1, oo);
                    }
                }
            }
        }
        int maxflow = dinic.solve(2*n+1, 2*n+2);
        int path_num = 0;
        for (int i = 0; i < dinic.en; i ++){
            if (i % 2 == 0)
                continue;
            if (dinic.arc[i].v == 2*n+1 || dinic.arc[i].u == 2*n+2)
                continue;
            if ((dinic.arc[i].u - 1 == dinic.arc[i].v) && (dinic.arc[i].u % 2 == 0))
                continue;
            if (dinic.arc[i].flow > 0){
                path[path_num].u = dinic.arc[i].v / 2;
                path[path_num].v = (dinic.arc[i].u + 1) / 2;
                path[path_num++].w = dinic.arc[i].flow;
            }
        }
        printf("%d %d
", maxflow, path_num);
        for (int i = 0; i < path_num; i ++){
            printf("%d %d %d
", path[i].u, path[i].v, path[i].w);
        }
    }
	return 0;
}

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原文地址：https://www.cnblogs.com/AbandonZHANG/p/4114039.html