洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows(01分数规划)

题意

题目链接

Sol

复习一下01分数规划

设\(a_i\)为点权，\(b_i\)为边权，我们要最大化\(\sum \frac{a_i}{b_i}\)。可以二分一个答案\(k\)，我们需要检查\(\sum \frac{a_i}{b_i} \geqslant k\)是否合法，移向之后变为\(\sum_{a_i} - k\sum_{b_i} \geqslant 0\)。把\(k * b_i\)加在出发点的点权上检查一下有没有负环就行了

#include<bits/stdc++.h> 
#define Pair pair<int, double>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define LL long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 4001, mod = 998244353, INF = 2e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
vector<Pair> v[MAXN];
double a[MAXN], dis[MAXN];
int vis[MAXN], times[MAXN];
bool SPFA(int S,  double k) {
    queue<int> q; q.push(S);
    for(int i = 1; i <= N; i++) vis[i] = 0, times[i] = 0, dis[i] = 0;
    times[S]++;
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = 0;
        for(auto &sta : v[p]) {
            int to = sta.fi; double w = sta.se;
            if(chmax(dis[to], dis[p] + a[p] - k * w)) {
                if(!vis[to]) q.push(to), vis[to] = 1, times[to]++;
                if(times[to] > N) return 1;
            }
        }
    }
    return 0;
}
bool check(double val) {
    for(int i = 1; i <= N; i++)
        if(SPFA(i, val)) return 1;
    return 0;
}
signed main() {
    N = read(); M = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read(), z = read();
        v[x].push_back({y, z});
    }
    double l = -1e9, r = 1e9;
    while(r - l > eps) {
        double mid = (l + r) / 2;
        if(check(mid)) l = mid;
        else r = mid;
    }
    printf("%.2lf", l);
    return 0;
}

洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows(01分数规划)

原文：https://www.cnblogs.com/zwfymqz/p/10479168.html