POJ - 3436 ACM Computer Factory 网络流

POJ-3436：http://poj.org/problem?id=3436

题意

　　组配计算机，每个机器的能力为x，只能处理一定条件的计算机，能输出特定的计算机配置。进去的要求有1，进来的计算机这个位子就要求为1，进去的要求有0，进来的计算机这个位子就要求为0.

思路

因为点上有容量限制，所以把每个点拆掉，连一条容量为这个机器的能力的边。源点向要求为0的机器连容量inf的边，把能完全组装好计算机的机器连向汇点。中间把符合条件的机器间连边，容量为inf；

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <iomanip>
#include   <cstdlib>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include    <cctype>
#include     <queue>
#include     <cmath>
#include      <list>
#include       <map>
#include       <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'
 
#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;
 
const ll mos = 0x7FFFFFFFLL;  //
const ll nmos = 0x80000000LL;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const double PI=acos(-1.0);
 
template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
            const int maxn = ;
            int p,n;
            struct node{
                int d;
                int in[maxn],out[maxn];
            }a[maxn];
 
            struct edge
            {
                int u,v,cap;
                edge(){}
                edge(int u,int v,int cap):
                u(u),v(v),cap(cap){}
            }es[];
            int tot,s,t;
            vector<int>tab[];
            int dis[],cur[];
            void addedge(int u,int v,int cap){
                tab[u].pb(tot);
                es[tot++] = edge(u,v,cap);
                tab[v].pb(tot);
                es[tot++] = edge(v,u,);
            }
 
            bool bfs(){
                queue<int>q;q.push(s);
                memset(dis,inf,sizeof(dis));
                dis[s] = ;
                while(!q.empty()){
                      int h = q.front();q.pop();
                      for(int i=; i<tab[h].size(); i++){
                            edge &e = es[tab[h][i]];
                            if(e.cap >  && dis[e.v] >= inf){
                                dis[e.v] = dis[h] + ;
                                q.push(e.v);
                            }
                    }
                }
                return dis[t] < inf;
            }
            int dfs(int x,int maxflow){
                if(x==t)return maxflow;
                for(int i=cur[x] ; i<tab[x].size(); i++){
                    cur[x] = i;
                    edge &e = es[tab[x][i]];
                    if(dis[e.v] == dis[x] +  && e.cap > ){
                        int flow = dfs(e.v, min(maxflow, e.cap));
                        if(flow){
                            e.cap -= flow;
                            es[tab[x][i] ^ ].cap += flow;
                            return flow;
                        }
                    }
              }
              return ;
            }
            int dinic(){
                int ans = ;
 
                while(bfs()){
                    int flow;
                    memset(cur,,sizeof(cur));
                    do{
                        flow = dfs(s,inf);
                        if(flow)ans += flow;
                    }while(flow);
                }
                return ans;
            }
int main(){
            scanf("%d%d", &p, &n);
            for(int i=; i<=n; i++){
                scanf("%d", &a[i].d);
                addedge(i,i+n,a[i].d);
                int t1 = ,t2 = ;
                for(int j=; j<=p; j++)scanf("%d",&a[i].in[j]), t1 += a[i].in[j]== ? :;
                for(int j=; j<=p; j++)scanf("%d",&a[i].out[j]),t2 +=  a[i].out[j]>? : ;
                if(t1 == ) addedge(, i, inf);
                if(t2 == p) addedge(i+n,n+n+,inf);
            }
            for(int i=; i<=n; i++){
                for(int j=; j<=n; j++){
                    if(i == j)continue;
                    int flag = ;
                    for(int t = ; t <= p; t++){
                        if(a[j].in[t] ==  && a[i].out[t] == )
                            flag = ;
                        if(a[j].in[t] ==  && a[i].out[t] == )
                            flag = ;
                    }
                    if(flag)addedge(i+n,j,inf);
                }
            }
            s = ,t = n+n+;
            printf("%d ", dinic());
            int rere = ;
            vector<p3>v;
            for(int i=; i<=n; i++){
                for(int j = ; j < tab[i+n].size(); j++){
                    edge e =  es[tab[i+n][j]];
                    if(e.cap < inf && i!= e.v&&e.v != t){
                       //printf("%d %d %d\n",i,e.v,inf - e.cap);
                        v.pb(p3(i,pii(e.v,inf - e.cap)));
                        rere ++;
                    }
                }
            }
            printf("%d\n", rere);
            for(int i=; i<rere; i++){
                printf("%d %d %d\n", v[i].fi,v[i].se.fi,v[i].se.se);
            }
            return ;
}

POJ 3436