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UVA 1161 - Objective: Berlin

题目链接

题意：给定一些航班，每个航班有人数，和起始终止时间，每次转机要花半小时，问限制时间内最多能有多少人从起始城市到终点城市

思路：以航班为结点建图，航班有容量限制所以进行拆点，然后两个航班如果终点和起点对上，并且时间满足就可以建边，然后源点连向起点为起始的航班，终点为终点的航班连向汇点（要在时间不超过时限的情况下），建好图跑一下最大流就可以了

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;

const int MAXNODE = 10005;
const int MAXEDGE = 1000005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
	}
};

struct Dinic {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool vis[MAXNODE];
	Type d[MAXNODE];
	int cur[MAXNODE];
	vector<int> cut;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}
	void add_Edge(int u, int v, Type cap) {
		edges[m] = Edge(u, v, cap, 0);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bfs() {
		memset(vis, false, sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (!vis[e.v] && e.cap > e.flow) {
					vis[e.v] = true;
					d[e.v] = d[u] + 1;
					Q.push(e.v);
				}
			}
		}
		return vis[t];
	}

	Type dfs(int u, Type a) {
		if (u == t || a == 0) return a;
		Type flow = 0, f;
		for (int &i = cur[u]; i != -1; i = next[i]) {
			Edge& e = edges[i];
			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i^1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}

	Type Maxflow(int s, int t) {
		this->s = s; this->t = t;
		Type flow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v])
				cut.push_back(i);
		}
	}
} gao;

const int N = 5005;

int n, hn, limit;
map<string, int> hash;

int get(string str) {
	if (!hash.count(str)) hash[str] = hn++;
	return hash[str];
}

int gett(string str) {
	int h = (str[0] - '0') * 10 + str[1] - '0';
	int m = (str[2] - '0') * 10 + str[3] - '0';
	return h * 60 + m;
}

struct HB {
	int u, v, c, s, t;
	void read() {
		string a, b;
		cin >> a >> b;
		u = get(a); v = get(b);
		cin >> c;
		cin >> a >> b;
		s = gett(a); t = gett(b);
	}
} h[N];

bool judge(HB a, HB b) {
	if (a.v != b.u) return false;
	if (a.t + 30 > b.s) return false;
	return true;
}

int main() {
	while (~scanf("%d", &n)) {
		string a, b;
		int s, t;
		hash.clear();
		hn = 0;
		cin >> a >> b;
		s = get(a); t = get(b); 
		cin >> a;
		limit = gett(a);
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			h[i].read();
		gao.init(n * 2 + 2);
		for (int i = 1; i <= n; i++) {
			if (h[i].u == s) gao.add_Edge(0, i, INF);
			if (h[i].v == t && h[i].t <= limit) gao.add_Edge(i + n, n * 2 + 1, INF);
			gao.add_Edge(i, i + n, h[i].c);
			for (int j = 1; j <= n; j++) {
				if (i == j) continue;
				if (judge(h[i], h[j])) gao.add_Edge(i + n, j, INF);
			}
		}
		printf("%d\n", gao.Maxflow(0, n * 2 + 1));
	}
	return 0;
}

UVA 1161 - Objective: Berlin(网络流)

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原文地址：http://blog.csdn.net/accelerator_/article/details/39578441