POJ - 3436 ACM Computer Factory(最大流，拆点)

题目链接
题目大意：要求经过每个点的流量不超过该点承载的流量上限，求最大流
思路：将a拆成a1和a2，a1作为输入端，a2作为输出端，a1->a2的边权为流量上限

#include <bits/stdc++.h>

using namespace std;
const int N = 2*100;
const int M = N*N;
int n,m,s,t,tot;
int head[N];
struct node{
    int u,nxt,cap,to,old;
}edge[M<<1];
int cur[N],deep[N],p[N],in[N][20],out[N][20];
bool check(int a,int b){   //检查是否可以a->b
    int i;
    for(i = 1;i <= m;i ++){
        if(in[b][i]==1&&out[a][i]==0) return 0;
        if(in[b][i]==0&&out[a][i]==1) return 0;
    }
    return 1;  //当时没写return 1本地居然过了，提交wa
}
void ae(int u,int v,int w){
    tot++;
    edge[tot].nxt = head[u];
    edge[tot].to = v;
    edge[tot].u = u;
    edge[tot].old = w;   //备份流量
    edge[tot].cap = w;
    head[u] = tot;
}
bool bfs(){
    memset(deep,-1,sizeof(deep));
    queue<int>q;
    q.push(s);
    deep[s] = 0;
    while(!q.empty()){
        int u = q.front();
        for(int i = head[u]; ~i;i = edge[i].nxt){
            int v = edge[i].to;
            if(deep[v]==-1&&edge[i].cap>0){
                q.push(v);
                deep[v] = deep[u]+1;
                if(v==t) return 1;
            }
        }
        q.pop();
    }
    return 0;
}
int dfs(int u,int f){
    int flow = 0,d;
    if(u==t||f==0) return f;
    for(int &i = cur[u]; ~i;i = edge[i].nxt){
        int v = edge[i].to;
        if(deep[v]>deep[u]&&edge[i].cap>0&&(d=dfs(v,min(f,edge[i].cap)))){
            edge[i].cap -= d;
            edge[i^1].cap += d;
            f -= d;
            flow += d;
            if(!f) break;
        }
    }
    if(flow==0) deep[u] = -1;
    return flow;
}
int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(cur, head, sizeof(head));
        ans += dfs(s,1e9);
    }
    return ans;
}
int hs(int x){
    if(x%2==1) return x/2+1;
    else return x/2;
}
int main()
{
    freopen("a.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin>>m>>n;
    s = 0;
    t = 2*n+1;
    memset(head,-1,sizeof(head));
    tot = -1;
    int i,j;
    for(i = 1;i <= n;i ++){
        cin>>p[2*i];
        ae(2*i-1,2*i,p[2*i]);    //拆点连边
        ae(2*i,2*i-1,0);
        for(j = 1;j <= m;j ++) cin>>in[2*i-1][j];
        for(j = 1;j <= m;j ++) cin>>out[2*i][j];
    }
    p[0] = 1e9;
    for(i = 1;i <= m;i ++) in[2*n+1][i] = 1;
    for(i = 0;i <= 2*n;i += 2)
        for(j = 1;j <= 2*n+1;j += 2){
            if(check(i,j)){
                if(i-1==j)continue; //防止重复建边
                ae(i,j,p[i]);   //建图,其实可以把p[i]改成1e9.....
                ae(j,i,0);
            }
    }
    cout<<dinic()<<' ';
    int use = 0,k;
    for(i = 2*n;i <= tot;i += 2){
        if(k = edge[i].old-edge[i].cap){
            int u = edge[i].u;
            int v = edge[i].to;
            if(u==0||v==2*n+1)continue;
            use ++;
        }
    }
    cout<<use<<endl;
    for(i = 2*n;i <= tot;i += 2){   //跳过拆点产生的路径
        if(k = edge[i].old-edge[i].cap){
            int u = edge[i].u;
            int v = edge[i].to;
            if(u==0||v==2*n+1)continue;   //跳过s和t
            cout<<hs(u)<<' '<<hs(v)<<' '<<k<<endl;
        }
    }
    return 0;
}