BOJ)11409 열혈강호 6

문제: icpc.me/11409

Maximum Cost Maximum Flow를 구하는 문제이다.

기존 MCMF 모델링에서 cost만 음수값으로 취해준 뒤 계산해주면 된다.

#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#define INF 987654321
using namespace std;
struct Edge {
    int v, cost, cap, rev;
    Edge(int v, int cost, int cap, int rev)
        : v(v), cost(cost), cap(cap), rev(rev){}
};
int n, m;
vector<vector<Edge>> vt;
vector<int> pv, pe;
void addEdge(int here, int there, int cost, int cap) {
    vt[here].emplace_back(there, cost, cap, vt[there].size());
    vt[there].emplace_back(here, -cost, 0, vt[here].size() - 1);
}
bool spfa(int src, int sink) {
    vector<int> v(n + m + 2, 0);
    vector<int> dist(n + m + 2, INF);
    queue<int> qu;
    qu.push(src);
    dist[src] = 0;
    v[src] = 1;
    while (qu.size()) {
        int here = qu.front();
        v[here] = 0;
        qu.pop();
        for (int i = 0; i < vt[here].size();i++) {
            int there = vt[here][i].v;
            int cap = vt[here][i].cap;
            if (cap&&dist[there] > dist[here] + vt[here][i].cost) {
                dist[there] = dist[here] + vt[here][i].cost;
                pv[there] = here;
                pe[there] = i;
                if (!v[there]) {
                    v[there] = 1;
                    qu.push(there);
                }
            }
        }
    }
    return dist[sink] != INF;
}
pair<int, int> solve(int src, int sink) {
    int flow = 0, cost = 0;
    while (spfa(src, sink)) {
        int minFlow = INF;
        for (int i = sink; i != src; i = pv[i]) {
            int prev = pv[i];
            int idx = pe[i];
            minFlow = min(minFlow, vt[prev][idx].cap);
        }
        for (int i = sink; i != src; i = pv[i]) {
            int prev = pv[i];
            int idx = pe[i];
            vt[prev][idx].cap -= minFlow;
            vt[i][vt[prev][idx].rev].cap += minFlow;
            cost -= vt[prev][idx].cost*minFlow;
        }
        flow += minFlow;
    }
    return{ flow,cost };
}
int main() {
    scanf("%d%d", &n, &m);
    vt.resize(n + m + 2);
    pv.assign(n + m + 2, -1);
    pe.assign(n + m + 2, -1);
    int src = n + m, sink = n + m + 1;
    for (int i = 0; i < n; i++) {
        int j;
        scanf("%d", &j);
        while (j--) {
            int a, b;
            scanf("%d%d", &a, &b);
            addEdge(i, a - 1 + n, -b, 1);
        }
    }
    for (int i = 0; i < n; i++)
        addEdge(src, i, 0, 1);
    for (int i = 0; i < m; i++)
        addEdge(i + n, sink, 0, 1);
    pair<int, int> res = solve(src, sink);
    printf("%d\n%d\n", res.first, res.second);
    return 0;
}