洛谷$P2057\ [SHOI2007]$ 善意的投票 网络流

正解:网络流

解题报告:

传送门!

$umm$看到每个人要么0要么1就考虑最小割呗,,,?

然后贡献有两种?一种是违背自己的意愿,一种是和朋友的意愿违背了

所以考虑开一排点分别表示每个人,然后$S$表示0$T$表示1每个人先分别连向自己的意愿

然后再每个人向朋友连边

然后就欧克了,,,?

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rc register char
#define rb register bool
#define t(i) edge[i].to
#define fy(i) edge[i].fy
#define w(i) edge[i].wei
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define e(i,x) for(ri i=head[x];~i;i=edge[i].nxt)

const int N=+,inf=1e9;
int ed_cnt=-,head[N],dep[N],S,T,cur[N],n,m;
struct ed{int to,nxt,wei;}edge[N*N];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch<'' || ch>''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(''<=ch && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void ad(ri x,ri y,ri z,ri tmp)
{edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],tmp};head[x]=ed_cnt;}
il bool bfs()
{
    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;
    while(!Q.empty())
    {ri nw=Q.front();Q.pop();e(i,nw)if(w(i) && !dep[t(i)]){dep[t(i)]=dep[nw]+,Q.push(t(i));if(t(i)==T)return ;}}
    return dep[T];
}
il int dfs(ri nw,ri flow)
{
    if(nw==T || !flow)return flow;ri ret=;
    for(ri &i=cur[nw];~i;i=edge[i].nxt)
        if(w(i) && dep[t(i)]==dep[nw]+)
        {ri tmp=dfs(t(i),min(flow,w(i)));flow-=tmp,w(i)-=tmp,ret+=tmp,w(i^)+=tmp;}
    if(!ret)dep[nw]=;
    return ret;
}
il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}

int main()
{
    memset(head,-,sizeof(head));n=read();m=read();S=;T=n+;
    rp(i,,n){read()?ad(i,S,,):ad(T,i,,);}rp(i,,m){ri x=read(),y=read();ad(x,y,,);}
    printf("%d\n",dinic());
    return ;
}