ZOJ 2676 Network Wars(网络流+分数规划)

传送门

题意：求无向图割集中平均边权最小的集合。

论文《最小割模型在信息学竞赛中的应用》原题。

分数规划。每一条边取上的代价为1。

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

const int N = 1e5 + 10;
const double EPS = 1e-5;
struct Edge { int v, next; double f; } edge[N];
int n, m, head[N], cnt, level[N], iter[N], x[N], y[N];
double z[N];
bool vis[N];
inline void add(int u, int v, double f) {
    edge[cnt].v = v; edge[cnt].f = f; edge[cnt].next = head[u]; head[u] = cnt++;
    edge[cnt].v = u; edge[cnt].f = f; edge[cnt].next = head[v]; head[v] = cnt++;
}
bool bfs() {
    for (int i = 1; i <= n; i++) iter[i] = head[i], level[i] = -1, vis[i] = 0;
    queue<int> que;
    que.push(1);
    level[1] = 0;   
    while (!que.empty()) {
        int u = que.front(); que.pop();
        for (int i = head[u]; ~i; i = edge[i].next) {
            int v = edge[i].v; double f = edge[i].f;
            if (level[v] < 0 && f) {
                level[v] = level[u] + 1;
                que.push(v);
            }
        }
    }
    return level[n] != -1;
}

double dfs(int u, double f) {
    if (u == n) return f;
    double flow = 0;
    for (int i = iter[u]; ~i; i = edge[i].next) {
        iter[u] = i;
        int v = edge[i].v;
        if (level[v] == level[u] + 1 && edge[i].f) {
            double w = dfs(v, min(f, edge[i].f));
            //if (fabs(w) < EPS) continue;
            f -= w;
            flow += w;
            edge[i].f -= w, edge[i^1].f += w;
            if (f <= 0) break;
        }
    }
    return flow;
}

double dinic() {
    double ans = 0;
    while (bfs()) ans += dfs(1, 1e8);
    return ans;
}

bool check(double mid) {
    memset(head, -1, sizeof(head));
    cnt = 0;
    double flow = 0;
    for (int i = 1; i <= m; i++) {
        if (z[i] > mid) add(x[i], y[i], z[i] - mid);
        else flow += z[i] - mid;   
    }
    flow += dinic();
    return flow >= EPS;
}

void get_ans(int u) {
    vis[u] = 1;
    for (int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].v;
        if (!vis[v] && edge[i].f > EPS) {
            get_ans(v);
        }
    }
}
int ans[N], tol;

int main() {
    int kase = 1;
    while (~scanf("%d%d", &n, &m)) {
        if (kase > 1) puts("");
        kase = 2;
        for (int i = 1; i <= m; i++) {
            x[i] = read();y[i] = read();
            int a = read();z[i] = (double)a;
        }
        double l = 0, r = 1e7;
        while (r - l > EPS) {
            double mid = (l + r) / 2.0;
            if (check(mid)) l = mid;
            else r = mid;
        }
        check(r);
        get_ans(1);
        tol = 0;
        for (int i = 1; i <= m; i++) {
            if (vis[x[i]] + vis[y[i]] == 1 || z[i] <= r) {
                ans[++tol] = i;     
            }
        }
        printf("%d\n", tol);
        for (int i = 1; i <= tol; i++) {
            if (i - 1) putchar(' ');
            printf("%d", ans[i]);
        }
        puts("");
    }
    return 0;
}

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