poj3436ACM Computer Factory(不用拆点+最大流)

题目请戳这里

题目大意：一台电脑有p个零件组成，有n个电脑组装机，每个组装机只能组装特定机器。即第i台组装机只能组装必含某些零件或者不能含有某些零件的半成品，通过这台组装机可以组装成另一台半成品/成品。每个组装机每个小时能组装的台数一定，给定n个组装机和组装条件，求每小时最多能装配多少台电脑，并输出效率最高时的一些可行方案。

题目分析：最大流。每个组装机为一个点，对于每个组装机，如果组装条件没有必须不能存在的零件，那么源点与之连一条边，边权为此组装机每小时最大组装量。如果某组装机组装成的机器为成品（每个零件都齐全了），与汇点连边，边权为此组装机每小时的工作量。然后对于每对组装机，如果i组装出的半成品符号j的组装条件，那么i与j建边，边权为i每小时的工作量。最后跑一遍最大流即可。输出方案就是在残余网络中一遍bfs，找出所有流量非0的边，输出即可。

PS：随便搜了一下，此题网上解法基本上都是人云亦云的拆点+网络流，却又讲不明白为什么要拆点。为什么要拆点？此题本来就是有向图，而且从源点建边的时候控制好流量，就不会流错的，为什么要拆点呢？

详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 55;const int M = 3000;const int inf = 0x3f3f3f3f;int p,n,num;struct node{    int to,next,c,f,pre;}arc[M];int tim[N],head[N],sta[N],que[N],cnt[N],dis[N],rpath[N];int ts[N][11],td[N][11];struct nd{    int u,v,f;}ans[M];void build(int s,int e,int cap){    arc[num].to = e;    arc[num].c = cap;    arc[num].f = 0;    arc[num].next = head[s];    head[s] = num ++;    arc[num - 1].pre = num;    arc[num].pre = num - 1;    arc[num].to = s;    arc[num].c = arc[num].f = 0;    arc[num].next = head[e];    head[e] = num ++;}bool isok(int x,int y){    int i;    for(i = 1;i <= p;i ++)    {        if((ts[y][i] == 1 && td[x][i] != 1) || (ts[y][i] == 0 && td[x][i] != 0))            return false;    }    return true;}void init(){    int i,j;    bool flag;    memset(head,-1,sizeof(head));    num = 0;    scanf("%d",&n);    for(i = 1;i <= n;i ++)    {        flag = true;        scanf("%d",&tim[i]);        for(j = 1;j <= p;j ++)        {            scanf("%d",&ts[i][j]);            if(ts[i][j] == 1)                flag = false;        }        if(flag)            build(0,i,tim[i]);        flag = true;        for(j = 1;j <= p;j ++)        {            scanf("%d",&td[i][j]);            if(td[i][j] != 1)                flag = false;        }        if(flag)            build(i,n + 1,tim[i]);    }    for(i = 1;i <= n;i ++)    {        for(j = 1;j <= n;j ++)        {            if(i == j)                continue;            if(isok(i,j))                build(i,j,tim[i]);        }    }}void re_Bfs(){    int i,front,rear;    for(i = 0;i <= n + 1;i ++)    {        dis[i] = n + 2;        cnt[i] = 0;    }    front = rear = 0;    que[rear ++] = n + 1;    cnt[0] = 1;    dis[n + 1] = 0;    while(front != rear)    {        int u = que[front ++];        for(i = head[u];i != -1;i = arc[i].next)        {            if(arc[arc[i].pre].c == 0 || dis[arc[i].to] < n + 2)                continue;            dis[arc[i].to] = dis[u] + 1;            cnt[dis[arc[i].to]] ++;            que[rear ++] = arc[i].to;        }    }}void ISAP(){    re_Bfs();    int i,u,maxflow = 0;    for(i = 0;i <= n + 1;i ++)        sta[i] = head[i];    u = 0;    while(dis[0] < n + 2)    {        if(u == n + 1)        {            int curflow = inf;            for(i = 0;i != n + 1;i = arc[sta[i]].to)                curflow = min(curflow,arc[sta[i]].c);            for(i = 0;i != n + 1;i = arc[sta[i]].to)            {                arc[sta[i]].c -= curflow;                arc[arc[sta[i]].pre].c += curflow;                arc[sta[i]].f += curflow;                arc[arc[sta[i]].pre].f -= curflow;            }            maxflow += curflow;            u = 0;        }        for(i = sta[u];i != -1;i = arc[i].next)            if(arc[i].c > 0 && dis[arc[i].to] + 1 == dis[u])                break;        if(i != -1)        {            sta[u] = i;            rpath[arc[i].to] = arc[i].pre;            u = arc[i].to;        }        else        {            if((-- cnt[dis[u]]) == 0)                break;            sta[u] = head[u];            int Min = n + 2;            for(i = sta[u];i != -1;i = arc[i].next)                if(arc[i].c > 0)                    Min = min(Min,dis[arc[i].to]);            dis[u] = Min + 1;            cnt[dis[u]] ++;            if(u != 0)                u = arc[rpath[u]].to;        }    }    printf("%d ",maxflow);}void solve(){    ISAP();    int ansnum = 0;    int i,front,rear;    front = rear = 0;    memset(cnt,0,sizeof(cnt));    cnt[0] = 1;    for(i = head[0];i != -1;i = arc[i].next)        if(arc[i].f > 0)        {            que[rear ++] = arc[i].to;            cnt[arc[i].to] = 1;        }    while(front != rear)    {        int u = que[front ++];        for(i = head[u];i != -1;i = arc[i].next)        {            if(arc[i].f > 0 && arc[i].to != n + 1)            {                ans[ansnum].u = u;                ans[ansnum].v = arc[i].to;                ans[ansnum ++].f = arc[i].f;                if(cnt[arc[i].to] == 0)                {                    que[rear ++] = arc[i].to;                    cnt[arc[i].to] = 1;                }            }        }    }    printf("%d\n",ansnum);    for(i = 0;i < ansnum;i ++)        printf("%d %d %d\n",ans[i].u,ans[i].v,ans[i].f);}int main(){    while(scanf("%d",&p) != EOF)    {        init();        solve();    }    return 0;}//180K16MS