洛谷P4926 [1007]倍杀测量者(差分约束)

题意

题目链接

sol

题目中的两个限制条件xian

\[a_i \geqslant (k_i - t)b_i\]

\[a_i(k_i + t) \geq b_i\]

我们需要让这两个至少有一个不满足

直接差分约束建边即可

这里要用到两个trick

1. 若某个变量有固定取值的时候我们可以构造两个等式\(c_i - 0 \leqslant x, c_i - 0 \geqslant x\)。

2. 乘法的大小判断可以取log变加法，因为\(y = log(x)\)也是个单调函数

#include<bits/stdc++.h> 
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long
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;}
using namespace std;
const int maxn = 4001, inf = 1e9;
const double eps = 1e-5;
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, k;
struct edge {
    int v, op;
    double k, w; 
};
vector<edge> v[maxn];
void addedge(int x, int y, double w, int opt, double k) {
    v[x].push_back({y, opt, k, w});
}
double dis[maxn];
bool vis[maxn];
int times[maxn];
bool spfa(double add) {
    queue<int> q; q.push(n + 1);
    for(int i = 0; i <= n; i++) dis[i] = -1e18, vis[i] = times[i] = 0;
    dis[n + 1] = 0; ++times[n + 1];
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = 0;
        for(auto &x : v[p]) {
            int opt = x.op, to = x.v; double k = x.k, w;
            if(opt == 0) w = x.w;
            else if(opt == 1) w = log2(k - add);
            else w = -log2(k + add);
            if(dis[to] < dis[p] + w) {
                dis[to] = dis[p] + w;
                if(!vis[to]) {
                    q.push(to);
                    vis[to] = 1;
                    ++times[to];
                    if(times[to] >= n + 1) return 0; 
                }
            }
        }
    }
    return 1;
}
signed main() {
    n = read(); m = read(); k = read();
    double l = 0, r = 10;
    for(int i = 1; i <= m; i++) {
        int opt = read(), x = read(), y = read(); double k = read();
        addedge(y, x, 0, opt, k);
        if(opt == 1) chmin(r, k);
    }
    for(int i = 1; i <= k; i++) {
        int c = read(); double x = read();
        addedge(0, c, log2(x), 0, 0);
        addedge(c, 0, -log2(x), 0, 0);
    }
    for(int i = 0; i <= n; i++) addedge(n + 1, i, 0, 0, 0);
    if(spfa(0))  return puts("-1"), 0;
    while(r - l > eps) {
        double mid = (r + l) / 2;
        if(spfa(mid)) r = mid;
        else l = mid;
    }
    printf("%lf", l);
    return 0;
}