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  • POJ

    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 << "
    ";
    #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 '
    '
    
    #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;  //2147483647
    const ll nmos = 0x80000000LL;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
    const double PI=acos(-1.0);
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',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 = 100;
                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[200009];
                int tot,s,t;
                vector<int>tab[10009];
                int dis[10009],cur[10009];
                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,0);
                }
    
                bool bfs(){
                    queue<int>q;q.push(s);
                    memset(dis,inf,sizeof(dis));
                    dis[s] = 0;
                    while(!q.empty()){
                          int h = q.front();q.pop();
                          for(int i=0; i<tab[h].size(); i++){
                                edge &e = es[tab[h][i]];
                                if(e.cap > 0 && dis[e.v] >= inf){
                                    dis[e.v] = dis[h] + 1;
                                    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] + 1 && e.cap > 0){
                            int flow = dfs(e.v, min(maxflow, e.cap));
                            if(flow){
                                e.cap -= flow;
                                es[tab[x][i] ^ 1].cap += flow;
                                return flow;
                            }
                        }
                  }
                  return 0;
                }
                int dinic(){
                    int ans = 0;
    
                    while(bfs()){
                        int flow;
                        memset(cur,0,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=1; i<=n; i++){
                    scanf("%d", &a[i].d);
                    addedge(i,i+n,a[i].d);
                    int t1 = 0,t2 = 0;
                    for(int j=1; j<=p; j++)scanf("%d",&a[i].in[j]), t1 += a[i].in[j]==1 ? 1:0;
                    for(int j=1; j<=p; j++)scanf("%d",&a[i].out[j]),t2 +=  a[i].out[j]>0?1 : 0;
                    if(t1 == 0) addedge(0, i, inf);
                    if(t2 == p) addedge(i+n,n+n+1,inf);
                }
                for(int i=1; i<=n; i++){
                    for(int j=1; j<=n; j++){
                        if(i == j)continue;
                        int flag = 1;
                        for(int t = 1; t <= p; t++){
                            if(a[j].in[t] == 1 && a[i].out[t] == 0)
                                flag = 0;
                            if(a[j].in[t] == 0 && a[i].out[t] == 1)
                                flag = 0;
                        }
                        if(flag)addedge(i+n,j,inf);
                    }
                }
                s = 0,t = n+n+1;
                printf("%d ", dinic());
                int rere = 0;
                vector<p3>v;
                for(int i=1; i<=n; i++){
                    for(int j = 0; 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
    ",i,e.v,inf - e.cap);
                            v.pb(p3(i,pii(e.v,inf - e.cap)));
                            rere ++;
                        }
                    }
                }
                printf("%d
    ", rere);
                for(int i=0; i<rere; i++){
                    printf("%d %d %d
    ", v[i].fi,v[i].se.fi,v[i].se.se);
                }
                return 0;
    }
    POJ 3436
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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/9749709.html
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