//Dinic是O(n2m)的。 #include<cstdio>#include<cstring>#include<cctype>#include<algorithm>using namespace std;#define rep(i,s,t) for(int i=s;i<=t;i++)#define dwn(i,s,t) for(int i=s;i>=t;i--)#define clr(x,c) memset(x,c,sizeof(x))int read(){    int x=0;char c=getchar();    while(!isdigit(c)) c=getchar();    while(isdigit(c)) x=x*10+c-‘0‘,c=getchar();    return x;}const int nmax=305;const int maxn=200000;const int inf=0x7f7f7f7f;struct edge{    int to,cap;edge *next,*rev;};edge es[maxn],*pt=es,*head[nmax];void add(int u,int v,int d){    pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;    pt->to=u;pt->cap=0;pt->next=head[v];head[v]=pt++;    head[u]->rev=head[v];head[v]->rev=head[u];}edge *cur[nmax],*p[nmax];int cnt[nmax],h[nmax];int maxflow(int s,int t,int n){    clr(cnt,0);cnt[0]=n;clr(h,0);    int flow=0,a=inf,x=s;edge *e;    while(h[s]<n){        for(e=cur[x];e;e=e->next) if(e->cap>0&&h[x]==h[e->to]+1) break;        if(e){            a=min(a,e->cap);p[e->to]=cur[x]=e;x=e->to;            if(x==t){                while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;                flow+=a;a=inf;            }        }else{            if(!--cnt[h[x]]) break;            h[x]=n;            for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1) h[x]=h[e->to]+1,cur[x]=e;            cnt[h[x]]++;            if(x!=s) x=p[x]->rev->to;        }    }    return flow;}int main(){    int n=read(),m=read(),u,v,s=0,t=n+1;    rep(i,1,n) {        u=read();        (!u)?add(i,t,1):add(s,i,1);    }    rep(i,1,m) u=read(),v=read(),add(u,v,1),add(v,u,1);    printf("%d\n",maxflow(s,t,t+1));    return 0;}

在第一个例子中，所有小朋友都投赞成票就能得到最优解

Day2