POJ-3436-ACM Computer Factory（網絡流）

鏈接：http://poj.org/problem?id=3436

拆點求最大流，並求出每邊的流量。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <string>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 105
#define MAXM 100005
int in_[55][12],out_[55][12],flow_[55];
int head[MAXN<<1],ind;
int deep[MAXN<<1];
int FLOW[MAXN<<1][MAXN<<1];
int all;
struct node
{
    int v, next,flow;
}edge[MAXM];
void add_edge(int u, int v, int flow)
{
    edge[ind].v=v;
    edge[ind].flow=flow;
    edge[ind].next=head[u];
    head[u]=ind++;
    edge[ind].v=u;
    edge[ind].flow=0;
    edge[ind].next=head[v];
    head[v]=ind++;
}
bool BFS(int S, int T)
{
    memset(deep,0,sizeof(deep));
    deep[S]=1;
    queue<int> q;
    q.push(S);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u]; i+1; i=edge[i].next)
        {
            int v= edge[i].v;
            if(!deep[v]&&edge[i].flow>0)
            {
                deep[v]=deep[u]+1;
                q.push(v);
                if(v==T)return 1;
            }
        }
    }
    return 0;
}
int DFS(int S, int T, int max_)
{
    if(S==T)return max_;
    int ans=0,f;
    for(int i=head[S]; i+1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(deep[S]+1==deep[v] && edge[i].flow>0)
        {
            f=DFS(v,T,min(max_-ans,edge[i].flow));
            ans+=f;
            FLOW[v][S]+=f;
            edge[i].flow-=f;
            edge[i^1].flow+=f;
        }
    }
    if(!ans)deep[S]=0;
    return ans;
}
int Dinic(int S, int T)
{
    int ans=0,t;
    while(BFS(S,T))
        while(t=DFS(S,T,INF))
        ans+=t;
    return ans;
}
bool link[120][120];
int a[MAXN],b[MAXN],c[MAXN];
int main()
{
    int n,m;
    int S=0,T=1;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        all=0;ind=0;
        memset(head,-1,sizeof(head));
        memset(FLOW,0,sizeof(FLOW));
        memset(link,0,sizeof(link));
        for(int i=1; i<=n; ++i)
        {
            int p=0;
            scanf("%d",&flow_[i]);
            for(int j=0; j<m; ++j)
            {
                scanf("%d",&in_[i][j]);
                if(in_[i][j]==0||in_[i][j]==2)p++;
            }
            if(p==m)
                add_edge(S,i<<1,INF);
            p=0;
            for(int j=0; j<m; ++j)
            {
                scanf("%d",&out_[i][j]);
                if(out_[i][j]==1)p++;
            }
            if(p==m)
                add_edge(i<<1^1,T,INF);
            add_edge(i<<1,i<<1^1,flow_[i]);
            for(int j=1; j<i; ++j)
            {
                p=1;
                int pp=1;
                for(int k=0; k<m; ++k)
                    if(out_[j][k]+in_[i][k]==1)
                    {
                        p=0;break;
                    }
                for(int k=0; k<m; ++k)
                    if(in_[j][k]+out_[i][k]==1)
                    {
                        pp=0;break;
                    }
                if(p)
                {
                    add_edge(j<<1^1,i<<1,INF);
                    link[j<<1^1][i<<1]=1;
                }
                if(pp)
                {
                    add_edge(i<<1^1,j<<1,INF);
                    link[i<<1^1][j<<1]=1;
                }
            }
        }
        int ans=Dinic(S,T);
        all=0;
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                if(i!=j)
                    if(link[i<<1^1][j<<1] && 0-FLOW[i<<1^1][j<<1]+FLOW[j<<1][i<<1^1])
                    {
                        a[all]=i;
                        b[all]=j;
                        c[all]=0-FLOW[i<<1^1][j<<1]+FLOW[j<<1][i<<1^1];
                        all++;
                       // printf("%d %d %d\n",i,j,0-FLOW[i<<1^1][j<<1]+FLOW[j<<1][i<<1^1]);
                    }
        printf("%d %d\n",ans,all);
        for(int i=0; i<all; ++i)
            printf("%d %d %d\n",a[i],b[i],c[i]);
    }
    return 0;
}

/*
3 4
15  0 0 0  0 1 0
10  0 0 0  0 1 1
30  0 1 2  1 1 1
3   0 2 1  1 1 1
3 5
5   0 0 0  0 1 0
100 0 1 0  1 0 1
3   0 1 0  1 1 0
1   1 0 1  1 1 0
300 1 1 2  1 1 1
2 2
100  0 0  1 0
200  0 1  1 1
*/