题解 P3199 【[HNOI2009]最小圈】

xyz32768 · 2017-10-22 11:00:22 · 题解

看到“平均值”，可以想到先二分圈的平均值的最小值，记当前答案为mid。

这时候，把所有的边权都减掉mid，这样判断原图中是否包含有平均值小于等于mid的圈，就转化成了判断新图中是否包含有负环。用SPFA或DFS判断负环即可。

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
    int res = 0; bool bo = 0; char c;
    while (((c = getchar()) < '0' || c > '9') && c != '-');
    if (c == '-') bo = 1; else res = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        res = (res << 3) + (res << 1) + (c - 48);
    return bo ? ~res + 1 : res;
}
const int N = 3005, M = 1e4 + 5; const double eps = 1e-9;
int n, m, ecnt, nxt[M], adj[N], go[M], w[M]; double val[M], dis[N];
bool Flag, vis[N];
void add_edge(int u, int v, int x) {
    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v; w[ecnt] = x;
}
void dfs(int u) {
    vis[u] = 1;
    for (int e = adj[u], v; e; e = nxt[e])
        if (dis[u] + val[e] < dis[v = go[e]]) {
            if (vis[v]) return (void) (Flag = 1);
            else {
                dis[v] = dis[u] + val[e];
                dfs(v);
            }
        }
    vis[u] = 0;
}
bool check(double x) {
    int i; memset(vis, 0, sizeof(vis));
    for (i = 1; i <= ecnt; i++) val[i] = 1.0 * w[i] - x;
    memset(dis, 0, sizeof(dis));
    for (i = 1; i <= n; i++) {
        Flag = 0;
        if (dfs(i), Flag) return 1;
    }
    return 0;
}
double solve() {
    double l = -1e5, r = 1e5;
    while (fabs(r - l) > eps) {
        double mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }
    return l;
}
int main() {
    int i, x, y, z; n = read(); m = read();
    for (i = 1; i <= m; i++) x = read(), y = read(),
        z = read(), add_edge(x, y, z);
    printf("%.8lf\n", solve());
    return 0;
}