BOJ)4716 풍선

문제: icpc.me/4716

모든 팀에게 풍선을 달아줄 때 풍선을 달기 위해 이동해야하는 최소 거리를 출력하는 문제이다.

그래프 모델링을 소스->방A,B -> N개의 팀-> 싱크  로 준 뒤 거리를 cost로 풍선을 capacity로 한 MCMF의 minimum cost를 출력하면 된다.

#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#define INF 987654321
using namespace std;
struct MCMF {
    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;
    vector<vector<Edge>> vt;
    vector<int> pv, pe;
    MCMF(int n) : n(n) {
        pv.assign(n + 1, -1);
        pe.assign(n + 1, -1);
        vt.resize(n + 1);
    }
    void addEdge(int s, int e, int cost, int cap) {
        vt[s].emplace_back(e, cost, cap, vt[e].size());
        vt[e].emplace_back(s, -cost, 0, vt[s].size() - 1);
    }
    bool spfa(int src, int sink) {
        vector<int> v(n + 1, 0);
        vector<int> dist(n + 1, 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 n, a, b, x, y, z;
int main() {
    while (scanf("%d%d%d", &n, &a, &b)) {
        if (!n&&!a&&!b)break;
        MCMF mcmf(n + 4);
        mcmf.addEdge(n + 1, n + 2, 0, a);
        mcmf.addEdge(n + 1, n + 3, 0, b);
        for (int i = 1; i <= n; i++) {
            scanf("%d%d%d", &x, &y, &z);
            mcmf.addEdge(n + 2, i, y, INF);
            mcmf.addEdge(n + 3, i, z, INF);
            mcmf.addEdge(i, n + 4, 0, x);
        }
        printf("%d\n", mcmf.solve(n + 1, n + 4).second);
    }
    return 0;
}