【PowerOJ1737&网络流24题】太空飞行计划问题（最小割）

题意：

思路：

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef long double ld;
  7 typedef pair<int,int> PII;
  8 typedef pair<ll,ll> Pll;
  9 typedef vector<int> VI;
 10 typedef vector<PII> VII;
 11 //typedef pair<ll,ll>P;
 12 #define N  200010
 13 #define M  200010
 14 #define INF 1e9
 15 #define fi first
 16 #define se second
 17 #define MP make_pair
 18 #define pb push_back
 19 #define pi acos(-1)
 20 #define mem(a,b) memset(a,b,sizeof(a))
 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 23 #define lowbit(x) x&(-x)
 24 #define Rand (rand()*(1<<16)+rand())
 25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 26 #define ls p<<1
 27 #define rs p<<1|1
 28
 29 const ll MOD=1e9+7,inv2=(MOD+1)/2;
 30       double eps=1e-6;
 31       int dx[4]={-1,1,0,0};
 32       int dy[4]={0,0,-1,1};
 33
 34 int head[N],vet[N],len[N],nxt[N],dis[N],s,S,T,tot;
 35
 36 int read()
 37 {
 38    int v=0,f=1;
 39    char c=getchar();
 40    while(c<48||57<c) {if(c==‘-‘) f=-1; c=getchar();}
 41    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 42    return v*f;
 43 }
 44
 45 void add(int a,int b,int c)
 46 {
 47     nxt[++tot]=head[a];
 48     vet[tot]=b;
 49     len[tot]=c;
 50     head[a]=tot;
 51
 52     nxt[++tot]=head[b];
 53     vet[tot]=a;
 54     len[tot]=0;
 55     head[b]=tot;
 56 }
 57
 58 bool bfs( )
 59 {
 60     queue<int>q;
 61     rep(i,1,s) dis[i]=-1;
 62     q.push(S),dis[S]=0;
 63     while(!q.empty())
 64     {
 65         int u=q.front();
 66         q.pop();
 67         int e=head[u];
 68         while(e)
 69         {
 70             int v=vet[e];
 71             if(len[e]>0&&dis[v]==-1)
 72             {
 73                 dis[v]=dis[u]+1;
 74                 q.push(v);
 75             }
 76             e=nxt[e];
 77         }
 78     }
 79     return dis[T]!=-1;
 80 }
 81
 82 int dfs(int u,int aug)
 83 {
 84     if(u==T) return aug;
 85     int e=head[u],val=0,flow=0;
 86     while(e)
 87     {
 88         int v=vet[e];
 89         if(len[e]>0&&dis[v]==dis[u]+1)
 90         {
 91             int t=dfs(v,min(len[e],aug));
 92             if(!t)
 93             {
 94                 e=nxt[e];
 95                 continue;
 96             }
 97             flow+=t;
 98             aug-=t;
 99             len[e]-=t;
100             len[e^1]+=t;
101             if(!aug) break;
102         }
103         e=nxt[e];
104     }
105     if(!flow) dis[u]=-1;
106     return flow;
107 }
108
109 int main()
110 {
111     int n=read(),m=read();
112     s=n+m;
113     S=++s,T=++s;
114     rep(i,1,s) head[i]=0;
115     tot=1;
116     int ans=0;
117     rep(i,1,n)
118     {
119         int x;
120         scanf("%d",&x);
121         add(S,i,x);
122         ans+=x;
123         while(getchar()==‘ ‘)
124         {
125             scanf("%d",&x);
126             add(i,x+n,INF);
127         }
128     }
129     rep(i,1,m)
130     {
131         int x=read();
132         add(i+n,T,x);
133     }
134     int tmp=0;
135     while(bfs()) tmp+=dfs(S,INF);
136     ans-=tmp;
137     rep(i,1,n)
138      if(dis[i]!=-1) printf("%d ",i);
139     printf("\n");
140     rep(i,1,m)
141      if(dis[i+n]!=-1) printf("%d ",i);
142     printf("\n");
143     printf("%d\n",ans);
144     return 0;
145 }

原文地址：https://www.cnblogs.com/myx12345/p/11743559.html