【POJ】【2987】Firing

网络流/最大权闭合子图

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　　胡伯涛论文里有讲……

　　sigh……细节处理太伤心了，先是count和ans输出弄反了，改过来顺序时又忘了必须先算出来ans！要是不执行一下Dinic的话count就无意义了……然后就是long long的问题……傻逼题白白WA了6次……sigh果然不能晚上搞……

 Source Code
 Problem:         User: sdfzyhy
 Memory: 2664K        Time: 344MS
 Language: G++        Result: Accepted

     Source Code

     //POJ 2987
     #include<vector>
     #include<cstdio>
     #include<cstring>
     #include<cstdlib>
     #include<iostream>
     #include<algorithm>
     #define rep(i,n) for(int i=0;i<n;++i)
     #define F(i,j,n) for(int i=j;i<=n;++i)
     #define D(i,j,n) for(int i=j;i>=n;--i)
     #define pb push_back
     using namespace std;
     inline int getint(){
         int v=,sign=; char ch=getchar();
         while(ch<''||ch>''){ if (ch=='-') sign=-; ch=getchar();}
         while(ch>=''&&ch<=''){ v=v*+ch-''; ch=getchar();}
         return v*sign;
     }
     const int N=,M=,INF=~0u>>;
     typedef long long LL;
     /******************tamplate*********************/
     int n,m;
     LL ans;
     struct edge{
         int from,to;
         LL v;
     };
     struct Net{
         edge E[M];
         int head[N],next[M],cnt;
         void add(int x,int y,LL v){
             E[++cnt]=(edge){x,y,v};
             next[cnt]=head[x]; head[x]=cnt;
             E[++cnt]=(edge){y,x,};
             next[cnt]=head[y]; head[y]=cnt;
         }
         int s,t,cur[N],d[N],Q[N];
         void init(){
             n=getint();m=getint();
             ans=;cnt=;
             s=;t=n+;
             LL x,y;
             F(i,,n){
                 x=getint();
                 if (x>=) {add(s,i,x);ans+=x;}
                 else add(i,t,-x);
             }
             F(i,,m){
                 x=getint(); y=getint();
                 add(x,y,INF);
             }
         }
         bool mklevel(){
             memset(d,-,sizeof d);
             d[s]=;
             int l=,r=-;
             Q[++r]=s;
             while(l<=r){
                 int x=Q[l++];
                 for(int i=head[x];i;i=next[i])
                     if (d[E[i].to]==- && E[i].v){
                         d[E[i].to]=d[x]+;
                         Q[++r]=E[i].to;
                     }
             }
             return d[t]!=-;
         }
         LL dfs(int x,LL a){
             if (x==t||a==) return a;
             LL flow=;
             for(int &i=cur[x];i && flow<a;i=next[i])
                 if (d[E[i].to]==d[x]+ && E[i].v){
                     LL f=dfs(E[i].to,min(a-flow,E[i].v));
                     E[i].v-=f;
                     E[i^].v+=f;
                     flow+=f;
                 }
             if (!flow) d[x]=-;
             return flow;
         }
         LL Dinic(){
             LL flow=;
             while(mklevel()){
                 F(i,s,t) cur[i]=head[i];
                 flow+=dfs(s,INF);
             }
             return flow;
         }
         bool vis[N];
         void dfs(int x){
             vis[x]=;
             for(int i=head[x];i;i=next[i])
                 if (!vis[E[i].to] && E[i].v) dfs(E[i].to);
         }
         void solve(){
             ans=ans-Dinic();
             int count=;
             memset(vis,,sizeof vis);
             dfs(s);
             F(i,,n) if (vis[i]) count++;
             printf("%d %lld\n",count,ans);
         }
     }G1;
     int main(){
     #ifndef ONLINE_JUDGE
         freopen("2987.in","r",stdin);
         freopen("2987.out","w",stdout);
     #endif
         G1.init();
         G1.solve();
         return ;
     }