P2762 太空飞行计划问题 最大权闭合子图

link:https://www.luogu.org/problemnew/show/P2762

题意

承担实验赚钱，但是要花去对应仪器的费用，仪器可能共用。
求最大的收益和对应的选择方案。

思路

这道题读入有点技巧，就是要自己判断换行
这道题和费用流关系不大，是最大权闭合子图，源点连接实验，容量为收益，实验向对应的仪器连接容量为inf的边，仪器向汇点连接容量为费用的边。
跑出最小割s，即最大流，然后用实验总收益 - s即可。
至于如何输出方案，即输出最大权闭合子图中的点。就是跑最后一次bfs的时候，有dis的点就是选中的点。
结合图形我们可以这么理解，从源点流到实验的边一定不能流满，要是满了，就说明这个做这个实验没有任何意义。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 1e4 + 9;
struct E {
    int u, v, w;
    int nxt;
} edge[maxn];
int gtot = 0, head[maxn];
void addedge(int u, int v, int w) {
    edge[gtot].u = u;
    edge[gtot].v = v;
    edge[gtot].w = w;
    edge[gtot].nxt = head[u];
    head[u] = gtot++;

    edge[gtot].u = v;
    edge[gtot].v = u;
    edge[gtot].w = 0;
    edge[gtot].nxt = head[v];
    head[v] = gtot++;
}
int dis[maxn], cur[maxn];
bool bfs(int s, int t) {
    rep(i, s, t) cur[i] = head[i];
    memset(dis, inf, sizeof(dis));
    dis[s] = 0;
    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v, w = edge[i].w;
            if (w > 0 && dis[v] > dis[u] + 1) {
                dis[v] = dis[u] + 1;
                que.push(v);
            }
        }
    }
    return dis[t] < inf;
}

int dfs(int u, int t, int maxflow) {
    if (u == t || maxflow == 0) return maxflow;

    for (int i = cur[u]; ~i; i = edge[i].nxt) {
        cur[u] = i;
        int v = edge[i].v, w = edge[i].w;
        if (w > 0 && dis[v] == dis[u] + 1) {
            int f = dfs(v, t, min(w, maxflow));
            if (f > 0) {
                edge[i].w -= f;
                edge[i ^ 1].w += f;
                return f;
            }
        }
    }
    return 0;
}

int dinic(int s, int t) {
    int flow = 0;
    while (bfs(s, t)) {
        while (int f = dfs(s, t, inf)) flow += f;
    }
    return flow;
}
char tools[10000];
int init[maxn];

int main() {
    memset(head, -1, sizeof(head));
    int n, m;
    scanf("%d%d", &n, &m);
    int s = 0, t = n + m + 1;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        int x;
        scanf("%d", &x);
        addedge(s, i, x);
        init[i] = x;
        sum += x;

        memset(tools, 0, sizeof tools);
        cin.getline(tools, 10000);
        int ulen = 0, tool;
        //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。

        while (sscanf(tools + ulen, "%d", &tool) == 1) {
            addedge(i, n + tool, inf);
            //  cout<<tool<<endl;
            if (tool == 0)
                ulen++;
            else {
                while (tool) {
                    tool /= 10;
                    ulen++;
                }
            }
            ulen++;
        }
    }
    for (int i = 1; i <= m; i++) {
        int x;
        scanf("%d", &x);
        addedge(i + n, t, x);
        init[i + n] = x;
    }
    int ans = sum - dinic(s, t);

    for (int i = 1; i <= n; i++) {
        if (dis[i] < inf) printf("%d ", i);
    }
    puts("");

    for (int i = 1; i <= m; i++) {
        if (dis[i + n] < inf) printf("%d ", i);
    }
    puts("");
    printf("%d\n", ans);
    return 0;
}