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相当于是

\[A_i \geqslant (K_i - T)B_i \]

\[A_i(K_i + T) \geq B_i \]

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;
}