POJ 3436 ACM Computer Factory (拆点+输出解)

忘是亡心i 2021-12-13 18:23 180阅读 0赞

【 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。 [![POJ 3436建图][POJ 3436]][POJ 3436]   【 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\n", maxflow, path_num);
            for (int i = 0; i < path_num; i ++){
                printf("%d %d %d\n", path[i].u, path[i].v, path[i].w);
            }
        }
    	return 0;
    }

转载于:https://www.cnblogs.com/AbandonZHANG/p/4114039.html

[POJ 3436]: /images/20211213/2a42583c86044564b796bfaf78909e95.png