nowcoder网络流勤奋的杨老师 最大权闭合子图

Posted 2021-08-15 kaka0010

原题链接：https://ac.nowcoder.com/acm/problem/15828

目录

• 题意
• 分析
• Code

题意

分析

这题算是最大权闭合子图的模板题了，有不了解的可以先看https://blog.csdn.net/can919/article/details/77603353

二选一问题转化为最小割的模型，然后再根据最大权闭合子图原理，如果选择一个点，那么前驱节点全部都选，因此向前驱节点连无限流量的边。然后再源点连正权点（流量为正权值），负权点连汇点（流量为负权的绝对值）就可以了。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const ll INF = 1e9;
const int N = 500 + 10;
const int M = 3e6 + 10;
const int MOD = 1e9 + 7;
int st, ed;
struct Maxflow {
    int h[N], cnt, maxflow, deep[N], cur[N];
    struct Edge {
        int to, next;
        ll cap;
    } e[M<<1];

    void init() {
        memset(h, -1, sizeof h);
        cnt = maxflow = 0;
    }
    void add(int u, int v, int cap) {
        e[cnt].to = v;
        e[cnt].cap = cap;
        e[cnt].next = h[u];
        h[u] = cnt++;

        e[cnt].to = u;
        e[cnt].cap = 0;
        e[cnt].next = h[v];
        h[v] = cnt++;

    }

    bool bfs() {
        for (int i = 0; i <= ed; i++) deep[i] = -1, cur[i] = h[i];
        queue<int> q; q.push(st); deep[st] = 0;
        while (q.size()) {
            int u = q.front();
            q.pop();
            for (int i = h[u]; ~i; i = e[i].next) {
                int v = e[i].to;
                if (e[i].cap && deep[v] == -1) {
                    deep[v] = deep[u] + 1;
                    q.push(v);
                }
            }
        }
        if (deep[ed] >= 0) return true;
        else return false;
    }

    ll dfs(int u, ll mx) {
        ll a;
        if (u == ed) return mx;
        for (int i = cur[u]; ~i; i = e[i].next) {
            cur[u] = i;//优化
            int v = e[i].to;
            if (e[i].cap && deep[v] == deep[u] + 1 && (a = dfs(v, min(e[i].cap, mx)))) {
                e[i].cap -= a;
                e[i ^ 1].cap += a;
                return a;
            }
        }
        return 0;
    }

    void dinic() {
        ll res;
        while (bfs()) {
            while (1) {
                res = dfs(st, INF);
                if (!res) break;
                maxflow += res;
            }
        }
    }
}mf;

int a[N];
void solve() {
    int n; cin >> n;
    for (int i = 1; i <= n; i++) {
        int x, y;
        cin >> x >> y;
        a[i] = x - y;
    }
    int u, v;
    st = n + 1, ed = n + 2;
    mf.init();
    while (cin >> u >> v) {
        mf.add(v, u, INF);
    }
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i] <= 0) mf.add(i, ed, -a[i]);
        else mf.add(st, i, a[i]), sum += a[i];
    }
    mf.dinic();
    cout << sum - mf.maxflow << endl;
}
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}