「JSOI2008」Blue Mary的旅行

传送门

Luogu

解题思路

分层图加网络流,有点像这题

可以证明最多不超过100天，所以才可以分层，不然图的规模会很大。

首先连源点汇点： \((s,1,INF), (n, t, INF)\)

以时间分层，每次把原图中的边 \((u, v, w)\) 改为一条 \((u_{day1}, v_{day2}, w)\) 的弧。

对于 \(u < n\)，连一条弧 \((u_{day1}, u_{day2}, INF)\)，以及一条 \((n_{day2}, n_{day1}, INF)\)。

然后每次在残量网络上加边增广直至最大流大于等于 \(T\)。

细节注意事项

每次增广时都要把流量累加

参考代码

#include <algorithm>

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <cstdio>

#include <cctype>

#include <cmath>

#include <ctime>

#include <queue>

#define rg register

using namespace std;

template < typename T > inline void read(T& s) {

	s = 0; int f = 0; char c = getchar();

	while (!isdigit(c)) f |= (c == '-'), c = getchar();

	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();

	s = f ? -s : s;

}

const int _ = 5010;

const int __ = 250010;

const int INF = 2147483647;

int tot = 1, head[_], nxt[__ << 1], ver[__ << 1], cap[__ << 1];

inline void Add_edge(int u, int v, int d)

{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, cap[tot] = d; }

inline void link(int u, int v, int d) { Add_edge(u, v, d), Add_edge(v, u, 0); }

int n, m, k, s, t, dep[_], cur[_];

struct node { int u, v, d; } x[2452];

inline int id(int u, int d) { return u + (d - 1) * n; }

inline int bfs() {

	static queue < int > Q;

	memset(dep, 0, sizeof dep);

	dep[s] = 1, Q.push(s);

	while (!Q.empty()) {

		int u = Q.front(); Q.pop();

		for (rg int i = head[u]; i; i = nxt[i]) {

			int v = ver[i];

			if (dep[v] == 0 && cap[i] > 0)

				dep[v] = dep[u] + 1, Q.push(v);

		}

	}

	return dep[t] > 0;

}

inline int dfs(int u, int flow) {

	if (u == t) return flow;

	for (rg int& i = cur[u]; i; i = nxt[i]) {

		int v = ver[i];

		if (dep[v] == dep[u] + 1 && cap[i] > 0) {

			int res = dfs(v, min(flow, cap[i]));

			if (res) { cap[i] -= res, cap[i ^ 1] += res; return res; }

		}

	}

	return 0;

}

inline int Dinic(int day) {

	int res = 0;

	while (bfs()) {

		cur[s] = head[s], cur[t] = head[t];

		for (rg int x = 1; x <= day; ++x)

			for (rg int i = 1; i <= n; ++i)

				cur[id(i, x)] = head[id(i, x)];

		while (int d = dfs(s, INF)) res += d;

	}

	return res;

}

int main() {

#ifndef ONLINE_JUDGE

	freopen("in.in", "r", stdin);

#endif

	read(n), read(m), read(k);

	for (rg int i = 1; i <= m; ++i)

		read(x[i].u), read(x[i].v), read(x[i].d);

	s = _ - 1, t = _ - 2;

	link(s, id(1, 1), INF);

	link(id(n, 1), t, INF);

	int res = 0;

	for (rg int d = 2; ; ++d) {

		for (rg int i = 1; i <= m; ++i)

			link(id(x[i].u, d - 1), id(x[i].v, d), x[i].d);

		for (rg int i = 1; i < n; ++i)

			link(id(i, d - 1), id(i, d), INF);

		link(id(n, d), id(n, d - 1), INF);

		res += Dinic(d);

		if (res >= k) { printf("%d\n", d - 1); return 0; }

	}

	return 0;

}

完结撒花 \(qwq\)

巴特西

「JSOI2008」Blue Mary的旅行

解题思路

细节注意事项

参考代码

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