ACM-ICPC 2018 沈阳赛区网络预赛 F. Fantastic Graph (有源汇上下界网络流板子)

板子题

给你一张二分图，一边n个点，另一边m个点，k条边，问你能不能通过删边，使得所有点的度数都在[L,R]

#include<bits/stdc++.h>
#define FIN freopen("input.txt","r",stdin)
#define ll long long
#define mod 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
const int maxn = 100005;
using namespace std;
int cnt,s,t,n,m,k;
int head[maxn],Next[maxn*20],to[maxn*20];
int flow[maxn*20],dep[maxn],cur[maxn],mp[405][405];
int dir[8][2]={1,2,1,-2,-1,2,-1,-2,-2,-1,-2,1,2,1,2,-1};
inline void add(int u,int v,int w){
    Next[cnt]=head[u];
    head[u]=cnt;
    to[cnt]=v;
    flow[cnt++]=w;

    Next[cnt]=head[v];
    head[v]=cnt;
    to[cnt]=u;
    flow[cnt++]=0;
}
int dfs(int u,int cost){
    if(u==t) return cost;
    for(int &i=cur[u];i!=-1;i=Next[i]){
        if(dep[to[i]]==dep[u]+1&&flow[i]>0){
           int dis=dfs(to[i],min(flow[i],cost));
            if(dis>0){
                flow[i]-=dis;
                flow[i^1]+=dis;
                return dis;
            }
        }
    }
    return 0;
}
int bfs(){
    queue<int> q;
    q.push(s);
    memset(dep,0,sizeof(dep));
    dep[s]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=Next[i]){
            if(flow[i]>0&&dep[to[i]]==0){
                dep[to[i]]=dep[u]+1;
                q.push(to[i]);
            }
        }
    }
    if(dep[t]>0) return 1;
    return 0;
}
int dicnic(){
    int ans=0;
    while(bfs()){
        int cost;
        for(int i=0;i<=t+2;i++){
            cur[i]=head[i];
        }
        while(cost=dfs(s,inf))
            ans+=cost;
    }
    return ans;
}
int p[200005];
void init(){
    memset(head,-1,sizeof(head));
    memset(p,0,sizeof(p));
    cnt=0;
}
int main(){
    int cas=0;
    while(~scanf("%d %d %d",&n,&m,&k)){
        init();
        int l,r;
        int ss=0; //源点
        int tt=n+m+1;//汇点
        s=n+m+2;//超级源点
        t=n+m+3;//超级汇点
        int sum=0;
        scanf("%d %d",&l,&r);
        for(int i=1;i<=k;i++){
            int u,v;
            scanf("%d %d",&u,&v);
            add(u,v+n,1);
        }
        for(int i=1;i<=n;i++){
            add(ss,i,r-l);
            p[i]+=l;
        }
        for(int i=1;i<=m;i++){
            add(i+n,tt,r-l);
            p[i+n]-=l;
        }
        for(int i=1;i<=n+m;i++){
            if(p[i]<0) add(i,t,-p[i]);
            else add(s,i,p[i]),sum+=p[i];
           // printf("%d ---> %d\n", i,p[i]);
        }
        add(tt,ss,inf);
        printf("Case %d: ",++cas);
        if(sum==dicnic())
            puts("Yes");
        else puts("No");
    }
}