luogu P1361 小M的作物

题目链接

luogu P1361 小M的作物

题解

源汇点为A,B

向种子连边,容量为价值,每个种子能与A或B联通,考虑最小割

用建边的总流量减去最小割就是答案

相同利益的时候新建节点,由额外利益构成群体向新建节点连边容量INF,新建节点向对应(源\汇)连对应额外利益的边

开始只新开了一个节点连了个双向边orz

代码

// luogu-judger-enable-o2
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::queue;
using std::min;
#define INF 0x7fffffff
const int maxn = 770000;
inline int read() {
    int x=0;char c=getchar();
    while(c<'0'||c>'9') c=getchar();
    while(c<='9'&&c>='0') x=x*10+c-'0',c=getchar();
    return x;
}
int n,S,T,sum,A[maxn],B[maxn];
struct node{
    int v,flow,next;
}edge[maxn];int head[maxn],num=1;
inline void add_edge(int u,int v,int flow) {
    edge[++num].v=v;edge[num].flow=flow,edge[num].next=head[u];head[u]=num;
    edge[++num].v=u;edge[num].flow=0;edge[num].next=head[v];head[v]=num;
}
int cur[maxn],lev[maxn];
bool spfa() {
    memset(lev,-1,sizeof lev);
    memcpy(cur,head,sizeof head);
    queue<int>que;
    que.push(S);lev[S]=0;
    while(!que.empty()) {
        int u=que.front();que.pop();
        for(int i=head[u];i;i=edge[i].next) {
            int v=edge[i].v;
            if(edge[i].flow>0&&lev[v]<0) {
                lev[v]=lev[u]+1;que.push(v);
            }
        }
    }
    if(lev[T]!=-1)return true;
    else return false;
}
int dfs(int now,int flow) {
    if(now==T)return flow;
    int rest=0,delta;
    for(int &i=cur[now];i;i=edge[i].next) {
        int v=edge[i].v;
        if(lev[v]==lev[now]+1&&edge[i].flow>0) {
            delta=dfs(v,min(flow-rest,edge[i].flow));
            if(delta) {
                edge[i].flow-=delta;
                edge[i^1].flow+=delta;
                rest+=delta;if(rest==flow)break;
            }
        }
    }
    if(rest==flow)lev[now]=-1;
    return rest;
}
int ans=0;
void dinic() {
    while(spfa())
         ans-=dfs(S,INF);
}
int main() {
    n=read();
    S=0,T=n+1;
    for(int a,i=1;i<=n;++i)
        add_edge(S,i,A[i]=read()),ans+=A[i];
    for(int a,i=1;i<=n;++i)
        add_edge(i,T,B[i]=read()),ans+=B[i];
    int m=read();
    for(int k,c1,c2,i=1;i<=m;++i) {
        k=read();c1=read(),c2=read();
        for(int a,j=1;j<=k;++j)
            add_edge(a=read(),T+i+m,INF),add_edge(T+i,a,INF);
        add_edge(S,T+i,c1),add_edge(T+i+m,T,c2);
        ans+=c1+c2;
    }
    dinic();
    printf("%d\n",ans);
    return 0;
}