[网络流24题] P2762 太空飞行计划

最大权闭合子图，需要方案
https://www.luogu.com.cn/problem/P2762

没啥好说的，裸的最大权闭合子图。
需要注意的是最后是否连通可以用 dep[v]判断

点击查看代码

#include <bits/stdc++.h>
#define endl '\n'
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0)
#define P pair<int, int>
typedef long long ll;
using namespace std;

const int N = 100 + 5;
const int M = N * N + 5;
const int INF = 0x3f3f3f3f;
struct edge {
    int v, w, to;
} e[M * 2];
int pre[N], cnt_edge, dep[N];
int S, T, z, head[N], sum;
int n, m, q[N], cur[N];
void add(int u, int v, int w) {
    e[cnt_edge] = {v, w, head[u]};
    head[u] = cnt_edge++;
    e[cnt_edge] = {u, 0, head[v]};
    head[v] = cnt_edge++;
}
bool bfs() {
    for (int i = 0; i <= T; i++) dep[i] = 0;
    dep[S] = 1;
    int l = 0, r = 1;
    q[r] = S;
    while (l < r) {
        int u = q[++l];
        for (int i = head[u]; i != -1; i = e[i].to) {
            int v = e[i].v;
            if (!dep[v] && e[i].w) dep[v] = dep[u] + 1, q[++r] = v;
        }
    }
    return dep[T];
}
int dfs(int u, int mi) {
    int res = 0;
    if (mi == 0 || u == T) return mi;
    for (int &i = cur[u]; i != -1; i = e[i].to) {
        int v = e[i].v;
        if (dep[u] + 1 == dep[v] && e[i].w) {
            int minn = dfs(v, min(mi - res, e[i].w));
            e[i].w -= minn;
            e[i ^ 1].w += minn;
            res += minn;
            if (res == mi) return res;
        }
    }
    if (res == 0) dep[u] = 0;
    return res;
}
int dinic() {
    ll res = 0;
    while (bfs()) {
        memcpy(cur, head, sizeof(head));
        //    cout<<res<<endl;
        res += dfs(S, INF);
    }
    return res;
}
int id(int x, int on) { return on * n + x; }
int vals[105];
int main() {
    int m;
    cin >> n >> m;
    memset(head, -1, sizeof head);
    S = n + m + 1, T = S + 1;
    int val;
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
        cin >> val;
        val = val > 0 ? val : -val;
        vals[i] = val;
        sum += val;
        add(S, id(i, 0), val);
        char tools[10000];
        memset(tools, 0, sizeof tools);
        cin.getline(tools, 10000);
        int ulen = 0, tool;
        while (sscanf(tools + ulen, "%d", &tool) == 1)
        //之前已经用scanf读完了赞助商同意支付该实验的费用
        {  // tool是该实验所需仪器的其中一个
            //这一行，你可以将读进来的编号进行储存、处理，如连边。
            // cout << "add: " << i << " " << tool << endl;
            add(id(i, 0), id(tool, 1), INF);
            if (tool == 0)
                ulen++;
            else {
                while (tool) {
                    tool /= 10;
                    ulen++;
                }
            }
            ulen++;
        }
    }

    for (int i = 1; i <= m; i++) {
        cin >> val;
        val = val > 0 ? val : -val;
        vals[i + n] = val;
        add(id(i, 1), T, val);
    }
    int ans = sum - dinic();
    // puts("-----------");
    for (int i = 1; i <= n; i++) {
        if (dep[i]) cout << i << " ";
    }
    cout << endl;
    for (int i = 1; i <= m; i++) {
        if (dep[i + n]) cout << i << " ";
    }
    cout << endl;
    cout << ans << endl;
    return 0;
}

放一下我最大权闭合子图的笔记

• 有一个有向图，每一个点都有一个权值（可以为正或负或0），选择一个权值和最大的子图，使得每个点的后继都在子图里面，这个子图就叫最大权闭合子图。
• 解决方法：
• 从源点s向每个正权点连一条容量为权值的边，每个负权点向汇点t连一条容量为权值的绝对值的边，原边容量全部为无限大。
• 求它的最小割，割掉后，与源点s连通的点构成最大权闭合子图，权值为正权值之和-最小割。

PS 用 \(Dinic\) 来写，是否与源点s连通可以用分层时的 \(dep\) 判断，其他写法可以用 \(dist\) 等

for (int i = 1; i <= n; i++) {
    if (dep[i]) cout << i << " ";//连通则输出
}

如：https://nanti.jisuanke.com/t/41096

点击查看代码

#include<iostream>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
#define int ll
using namespace std;
typedef long long ll;
const int M = 4000 + 5;
const int N = 8e2 + 5;
const int INF = 0x3f3f3f3f;
int n1,n2,tot;
struct node
{
    int v,w,to;
} edge[M*2];
int pre[N],cnt,dep[N];
int S,T,z,head[N],sum,id;
int n,m,q[N],cur[N];
void add(int u,int v,int w)
{
    edge[cnt]= {v,w,head[u]};
    head[u]=cnt++;
    edge[cnt]= {u,0,head[v]};
    head[v]=cnt++;
}
bool bfs()
{
    for(int i=0; i<=T; i++)
        dep[i]=0;
    dep[S]=1;
    int l=0,r=1;
    q[r]=S;
    while(l<r)
    {
        int u=q[++l];
        for(int i=head[u]; i!=-1; i=edge[i].to)
        {
            int v=edge[i].v;
            if(!dep[v]&&edge[i].w)
                dep[v]=dep[u]+1,q[++r]=v;
        }
    }
    return dep[T];
}
int dfs(int u,int mi)
{
    int res=0;
    if(mi==0||u==T)
        return mi;
    for(int &i=cur[u]; i!=-1; i=edge[i].to)
    {
        int v=edge[i].v;
        if(dep[u]+1==dep[v]&&edge[i].w)
        {
            int minn=dfs(v,min(mi-res,edge[i].w));
            edge[i].w-=minn;
            edge[i^1].w+=minn;
            res+=minn;
            if(res==mi)
                return res;
        }
    }
    if(res==0) dep[u]=0;
    return res;
}
int dinic()
{
    ll res=0;
    while(bfs())
    {
        memcpy(cur,head,sizeof(head));
//        cout<<res<<endl;
        res+=dfs(S,INF);
    }
    return res;
}

signed main(){
	int t;cin >> t;
	while(t--){
		cin >> n1 >> n2;
		S = 0;T = n1 + n2 + 1;cnt= 0;
		ll totp = 0;
		for(int i = 0; i <= n1 + n2 + 1; i++) head[i] = -1;
		
		for(int i = 1; i <= n1; i++){
			int w;cin >> w;
			totp += w;
			add(S,i,w);
		}
		for(int i = 1; i <= n2; i++){
			int w;cin >> w;
			add(i+n1,T,w);
		}
		
		int m1,m2,v;
		for(int u = 1; u <= n1; u++){
			cin >> m1 >> m2;
			for(int i = 0; i < m1; i++){
				cin >> v;
				add(u,v+n1,INF);
			}
			for(int i = 0; i < m2; i++){
				cin >> v;
				add(u,v,INF);
			}
		}
		int ans = dinic();
		cout<<totp - ans<<endl;
	}
}