网络流24题 P2762 太空飞行计划问题

题目链接

思路

问题模型：最大权闭合图

转化模型：网络最小割

这道题是网络流中一个比较重要的模型：最大权闭合图转最大流

建立超级源点\(S\)和超级汇点\(T\)，然后每个实验连一条从\(S\)到实验，流量为实验收益的边，每个仪器连一条从仪器到\(T\), 流量为仪器耗费的边，然后需要的仪器就连一条从实验到仪器流量为\(inf\)（无穷大）的边，因为实验到仪器的边的流量为正无穷，所以最小割一定不会在上面，根据最大流最小割定理，最大流就等于最小割，我们按照以上所说建图，求出最大流，之后用实验利益的总和减去最大流，得出的就是最大净收益

最大流算法我用的是\(\text{Dinic}\)算法，因为这样方便输出，为什么？因为如果\(d[i]\)不为\(0\)就说明它一定用过，这样就能方便输出啦~

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 1e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
    char c = getchar(); int x = 0, f = 1;
    for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return x * f;
}

int m, n, cnt, opt, S, T, ans, head[A], d[A], q[A];
struct node { int from, to, nxt, val; } e[A];

inline void add(int from, int to, int val) {
    e[cnt].to = to;
    e[cnt].val = val;
    e[cnt].nxt = head[from];
    head[from] = cnt++;
}

inline bool makelevel(int s, int t) {
    memset(d, 0, sizeof(d));
    memset(q, 0, sizeof(q));
    int l = 0, r = 0;
    d[s] = 1; q[r++] = s;
    while (l < r) {
        int x = q[l++];
        if (x == t) return true;
        for (int i = head[x]; i != -1; i = e[i].nxt) {
            int to = e[i].to;
            if (d[to] == 0 && e[i].val > 0) {
                d[to] = d[x] + 1;
                q[r++] = e[i].to;
            }
        }
    }
    return false;
}

int dfs(int x, int flow, int t) {
    if (x == t) return flow;
    int sum = 0;
    for (int i = head[x]; i != -1; i = e[i].nxt) {
        int to = e[i].to;
        if (d[to] == d[x] + 1 && e[i].val > 0) {
            int tmp = dfs(to, min(flow - sum, e[i].val), t);
            e[i].val -= tmp, e[i ^ 1].val += tmp;
            sum += tmp;
            if (sum == flow) return sum;
        }
    }
    return sum;
}

int main() {
    memset(head, -1, sizeof(head));
    m = read(), n = read();
    int S = 0, T = 555;
    int w, tot = 0, x;
    for (int i = 1; i <= m; i++) {
        scanf("%d", &w), tot += w;
        add(S, i, w), add(i, S, 0);
        while (getchar() == ' ') {
            scanf("%d", &x);
            add(i, x + m, inf);
            add(x + m, i, 0);
        }
    }
    for (int i = 1; i <= n; i++) {
        x = read();
        add(i + m, T, x), add(T, i + m, 0);
    }
    while (makelevel(S, T)) ans += dfs(S, inf, T);
    ans = tot - ans;
    for (int i = 1; i <= m; i++) if (d[i]) cout << i << ' '; puts("");
    for (int i = 1; i <= n; i++) if (d[i + m]) cout << i << ' '; puts("");
    cout << ans << '\n';
    return 0;
}

原文地址：https://www.cnblogs.com/loceaner/p/12310836.html