[SDOI2009]晨跑

这道题是一个很裸的费用流题目。

一般点之过一次的套路就是拆点连边。

其他的模拟下就水过去了。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define INF (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 200 + 10, maxm = 20000 + 10;

struct Edge {
    int u, v, c, f, w, pre;
};

int mincost = INF;

struct Graph {
    Edge ed[maxm << 2];
    int n, m, G[maxn << 1];
    void init(int n) {
        this->n = n;
        m = 1;
        memset(G, 0, sizeof(G));
    }
    void add(int u, int v, int c, int w) {
        ed[++m] = (Edge){u, v, c, 0, w, G[u]};
        G[u] = m;
        ed[++m] = (Edge){v, u, 0, 0, -w, G[v]};
        G[v] = m;
    }
    int a[maxn << 1], d[maxn << 1], p[maxn << 1], inq[maxn << 1];
    int s, t;
    queue<int> Q;
    bool spfa(int &flow, int &cost) {
        memset(inq, 0, sizeof(inq));
        rep(i, 1, n) d[i] = INF;
        d[s] = 0;
        a[s] = INF;
        Q.push(s); inq[s] = 1;
        while (!Q.empty()) {
            int u = Q.front(); Q.pop(); inq[u] = 0;
            for (register int i = G[u]; i; i = ed[i].pre) {
                Edge &e = ed[i];
                if (e.f < e.c && d[u] + e.w < d[e.v]) {
                    d[e.v] = d[u] + e.w;
                    a[e.v] = min(a[u], e.c-e.f);
                    p[e.v] = i;
                    if (!inq[e.v]) Q.push(e.v), inq[e.v] = 1;
                }
            }
        }
        if (d[t] == INF) return false;
        flow += a[t];
        cost += a[t] * d[t];
        for (register int i = p[t]; i; i = p[ed[i].u]) {
            ed[i].f += a[t];
            ed[i^1].f -= a[t];
        }
        return true;
    }
    int mcmf(int s, int t, int &cost) {
        this->s = s, this->t = t;
        int flow = 0; cost = 0;
        while (spfa(flow, cost));
        return flow;
    }
} G;

int n, m;

int main() {
    n = read(), m = read(); G.init(n*2);
    rep(i, 1, m) {
        int u = read(), v = read(), w = read();
        G.add(u+n, v, 1, w);
    }
    rep(i, 2, n-1) G.add(i, i+n, 1, 0);
    G.add(n, n*2, INF, 0);
    G.add(1, n+1, INF, 0);
    int cost;
    printf("%d", G.mcmf(1, n*2, cost));
    printf(" %d", cost);
    return 0;
}