01分数规划与图论

1.什么是01分数规划?

0/1分数规划模型是指,给定整数和,求一组解使得下列式子最大化:.

2.解决方法

我们最优解是L,即,也就是.

我们令,做出函数图像

不同的直线,对于不同的选择方案,与此同时,直线横截距就是答案,我们要求出最大的横截距.

我们可以随机产生一组答案,然后判断此时 是否大于0,如果大于0,那么根据图像,我们得把解增大,如果小于0我们得把解变小,如果等于0,则无法判断.

这个过程可以通过二分实现.

那么我们怎么求呢?具体的方式看题目,接下来,我们将会解决两道题目来熟悉这个知识点.

3.例题与实战

例题一:AcWing361 题目传送门

思路:

我们要求的最大值,a是点权,b是边权,我们二分mid,时,答案需要调大一点,时,答案需要调小一些,后面的式子可以转化为,此时我们把点权转到边权上,边权为a_i - b_i * mid,此时跑spfa判断有没有负环即可.

代码:

#include 
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
#define double long double
#define ull unsigned long long 
#define PII pair 
#define PDI pair 
#define PDD pair 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 1010, M = 2e6 + 10;
const double eps = 1e-5;
int n, m, cnt[N], h[N], v[M], to[M], tot;
double w[M], f[N], dis[N];
bool vis[N];
void add(int a, int b, double c) {
	w[++tot] = c, v[tot] = b, to[tot] = h[a], h[a] = tot;
}
bool check(double t) {
	queue<int> q;
	for (int i = 1; i <= n; ++i) dis[i] = 0, cnt[i] = 0, vis[i] = 0, q.emplace(i);
	while (q.size()) {
		int x = q.front(); q.pop();
		vis[x] = 0;
		for (int i = h[x]; i; i = to[i]) {
			int y = v[i];
			double z = t * w[i] - f[x];
			if (dis[y] > dis[x] + z) {
				dis[y] = dis[x] + z;
				cnt[y] = cnt[x] + 1;
				if (cnt[y] >= n) return true;
				if (!vis[y]) q.emplace(y), vis[y] = 1;
			}
		}
	}
	return false;
}
signed main() {
	IOS;
	point(2);
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) cin >> f[i];
	for (int i = 1; i <= m; ++i) {
		int a, b;
		double c;
		cin >> a >> b >> c;
		add(a, b, c);
	}
	double l = 0, r = 1e9;
	while (r - l > eps) {
		double mid = (l + r) / 2.0;
		if (check(mid)) l = mid;
		else r = mid;
	}
	cout << l << "\n";
}

例题二.AcWing348 题目传送门

思路:

二分答案,然后建新图,边权为比率,然后跑最小生成树即可.

代码:

#include 
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
// #define double long double
#define ull unsigned long long 
#define PII pair 
#define PDI pair 
#define PDD pair 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 2e6 + 10, M = 1010;
const double eps = 1e-6;
int n;
double x[N], y[N], z[N], w[M][M], dis[N];
bool vis[N];
double calc(int a, int b) {
	return sqrt((x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]));
}
bool check(double x) {
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j)
			w[i][j] = w[j][i] = abs(z[i] - z[j]) - x * calc(i, j);
	for (int i = 1; i <= n; ++i) dis[i] = 1e18, vis[i] = 0;
	dis[1] = 0;
	for (int i = 1; i < n; ++i) {
		int x = 0;
		for (int j = 1; j <= n; ++j) 
			if (!vis[j] && (x == 0 || dis[j] < dis[x])) x = j;
		vis[x] = 1;
		for (int y = 1; y <= n; ++y)
			if (!vis[y]) dis[y] = min(dis[y], w[x][y]);
	}
	double ans = 0;
	for (int i = 2; i <= n; ++i) ans += dis[i];
	return ans >= 0;
}
signed main() {
	IOS;
	point(3);
	while (cin >> n, n) {
		for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i] >> z[i];
		double l = 0, r = 1000000000.0;
		while (r - l > eps) {
			double mid = (l + r) / 2.0;
			if (check(mid)) l = mid;
			else r = mid;
		}
		cout << l << "\n";
	}
}