[BZOJ 2285] [SDOI 2011] 保密

Description

传送门

Solution

这道题的最大难点在于读懂题意（雾

分数规划求出 \(n\) 到 \(1\cdots n_1\) 每个点的最小 \(\sum\frac{t_i}{s_i}\)，然后转换成最小点权覆盖问题，最小点权覆盖 = 最大匹配数。

Code

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>

const double eps = 1e-8, INF = 1e9;
struct Edge1 { int v, nxt; double f, c; } e[80325];
struct Edge2 { int v, nxt, t, s; double w; } g[100005];
int n, m, nn, mm, hd[702], head[165], tot, S, T, cur[165], d[165], cnt;

int read() {
    int x = 0; char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    return x;
}
void add(int u, int v, int t, int s) {
    g[++tot].nxt = hd[u], hd[u] = tot, g[tot].v = v, g[tot].t = t, g[tot].s = s;
}
void adde(int u, int v, double c) {
    e[++tot].nxt = head[u], head[u] = tot, e[tot].v = v, e[tot].f = 0, e[tot].c = c;
    e[++tot].nxt = head[v], head[v] = tot, e[tot].v = u, e[tot].f = 0, e[tot].c = 0;
}
bool bfs() {
    bool vis[165] = {}; std::queue<int> q; vis[S] = 1, q.push(S);
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = head[u]; i; i = e[i].nxt)
            if (!vis[e[i].v] && e[i].c > e[i].f + eps)
                d[e[i].v] = d[u] + 1, vis[e[i].v] = 1, q.push(e[i].v);
    }
    return vis[T];
}
double dfs(int u, double a) {
    if (u == T || a < eps) return a;
    double flow = 0, x;
    for (int &i = cur[u]; i; i = e[i].nxt)
        if (d[u] + 1 == d[e[i].v] && (x = dfs(e[i].v, std::min(a, e[i].c - e[i].f))) > eps) {
            e[i].f += x, e[i ^ 1].f -= x, flow += x, a -= x;
            if (a < eps) break;
        }
    return flow;
}
double dinic() {
    double flow = 0;
    while (bfs()) memcpy(cur, head, sizeof head), flow += dfs(S, INF);
    return flow;
}
bool check(int x, double mid) {
    int d[702] = {}; double f[702] = {}; std::queue<int> q;
    for (int i = 1; i <= m; ++i) g[i].w = g[i].t - mid * g[i].s, ++d[g[i].v];
    for (int i = 1; i <= n; ++i) { if (!d[i]) q.push(i); f[i] = INF; }
    f[n] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = hd[u]; i; i = g[i].nxt) {
            f[g[i].v] = std::min(f[g[i].v], f[u] + g[i].w);
            if (--d[g[i].v] == 0) q.push(g[i].v);
        }
    }
    return f[x] > eps;
}
bool bfs2() {
    int cnt = 0, vis[702] = {}; std::queue<int> q;
    q.push(n), vis[n] = 1, cnt += (n == nn);
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = hd[u]; i; i = g[i].nxt) if (!vis[e[i].v]) {
            vis[e[i].v] = 1, q.push(e[i].v);
            if ((cnt += (e[i].v <= nn)) == nn) return false;
        }
    }
    return cnt < nn;
}
int main() {
    n = read(), m = read();
    for (int i = 1, u, v, t, s; i <= m; ++i)
        u = read(), v = read(), t = read(), s = read(), add(u, v, t, s);
    tot = 1, mm = read(), nn = read(), S = nn + 1, T = S + 1;
    for (int i = 1, u, v; i <= mm; ++i) u = read(), v = read(), adde(u, v, INF);
    if (bfs2()) { puts("-1"); return 0; }
    for (int i = 1; i <= nn; ++i) {
        double l = 0, r = 10, mid;
        while (l + eps < r) {
            mid = (l + r) / 2;
            if (check(i, mid)) l = mid; else r = mid;
        }
        if (i & 1) adde(S, i, l); else adde(i, T, l);
    }
    printf("%.1lf\n", dinic());
    return 0;
}

标签：SDOI,int,double,flow,2285,eps,vis,while,2011
来源： https://www.cnblogs.com/fly-in-milkyway/p/10419846.html