BZOJ4155 : [Ipsc2015]Humble Captains

第一问最小割，第二问：

设du[i]表示i点的度数，则要最小化$\frac{|1集合的du[i]之和-2集合的du[i]之和|}{2}$，

压位01背包即可。

#include<cstdio>
#include<bitset>
using namespace std;
const int N=40010,inf=~0U>>2;
struct edge{int t,f;edge*nxt,*pair;}*g[N],*d[N],pool[240000],*cur=pool;
int Case,n,m,cnt,i,x,y,S,T,h[N],gap[N],maxflow,du[N],ans;bitset<19910>f;
inline int abs(int x){return x>0?x:-x;}
inline int min(int x,int y){return x<y?x:y;}
inline void add(int s,int t,int f){edge*p=cur++;p->t=t;p->f=f;p->nxt=g[s];g[s]=p;p=cur++;p->t=s;p->f=0;p->nxt=g[t];g[t]=p;g[s]->pair=g[t];g[t]->pair=g[s];
}
int sap(int v,int flow){if(v==T)return flow;int rec=0;for(edge*p=d[v];p;p=p->nxt)if(h[v]==h[p->t]+1&&p->f){int ret=sap(p->t,min(flow-rec,p->f));p->f-=ret;p->pair->f+=ret;d[v]=p;if((rec+=ret)==flow)return flow;}if(!(--gap[h[v]]))h[S]=T;gap[++h[v]]++;d[v]=g[v];return rec;
}
int main(){for(scanf("%d",&Case);Case--;printf("%d\n",ans)){scanf("%d%d",&n,&m),S=n+m+m+1,T=S+1,cnt=n;for(i=1;i<=m;i++){scanf("%d%d",&x,&y);du[x]++,du[y]++;add(S,++cnt,1),add(++cnt,T,1);add(cnt-1,x,inf),add(x,cnt,inf);add(cnt-1,y,inf),add(y,cnt,inf);}add(S,1,inf),add(2,T,inf);for(gap[maxflow=0]=T,i=1;i<=T;i++)d[i]=g[i];while(h[S]<T)maxflow+=sap(S,inf);printf("%d ",m*2-maxflow);for(cur=pool,i=0;i<=T;i++)g[i]=d[i]=NULL,h[i]=gap[i]=0;for(f.reset(),f[0]=1,i=3;i<=n;i++)f|=f<<du[i];for(ans=inf,i=0;i<=m;i++)if(f[i])ans=min(ans,abs(du[1]+i-m));for(i=1;i<=n;i++)du[i]=0;}return 0;
}

转载于:https://www.cnblogs.com/clrs97/p/4608833.html

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