G - Gasoline Gym - 101908G

传送门：QAQ

题意：给你一些接受点和提供点，然后给你一些接受点到提供点的边和所需时间，求最短时间（同时运输）将所有接受点装满，输出所需时间。

思路：二分时间，然后把满足时间的点加入边，跑网络流即可。

附上代码：

#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 21000;
#define inf 0x3f3f3f3f
int ax[1100];
int bn[1100];
struct inst {
	int x, y, w1;
};
inst ww[21000];
int n, m;
int n1, m1, k;
int sum1;
int sum2;
struct Edge {
	Edge() {}
	Edge(int from, int to, int cap, int flow) :from(from), to(to), cap(cap), flow(flow) {}
	int from, to, cap, flow;
};
struct Dinic {
	int n, m, s, t;
	vector<Edge>edges;
	vector<int>G[maxn];
	bool vis[maxn];
	int d[maxn];
	int cur[maxn];
	void init(int n) {
		this->n = n;
		for (int i = 0; i <= n; i++) G[i].clear();
		edges.clear();
	}
	void AddEdge(int from, int to, int cap) {
		edges.push_back(Edge(from, to, cap, 0));
		edges.push_back(Edge(to, from, 0, 0));
		m = edges.size();
		G[from].push_back(m - 2);
		G[to].push_back(m - 1);
	}
	bool BFS() {
		memset(vis, 0, sizeof(vis));
		queue<int>Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int x = Q.front();
			Q.pop();
			for (int i = 0; i < G[x].size(); i++) {
				Edge& e = edges[G[x][i]];
				if (!vis[e.to] && e.cap > e.flow) {
					vis[e.to] = true;
					d[e.to] = d[x] + 1;
					Q.push(e.to); 
				}
			}
		}
		return vis[t];
	}
	int DFS(int x, int a) {
		if (x == t || a == 0) return a;
		int flow=0,f;
		for (int& i = cur[x]; i < G[x].size(); i++) {
			Edge& e = edges[G[x][i]];
			if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[G[x][i] ^ 1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}
	int Maxflow(int s, int t) {
  		this->s = s, this->t = t;
		int flow = 0;
		while (BFS()) {
			memset(cur, 0, sizeof(cur));
			flow += DFS(s, inf);
                                                    		}
		return flow;
	}
}DC;
int check(int mid) {
	DC.init(2100);
	int S = 0;
	int T = n1 + m1 + 1;
	for (int i = 1; i <= m1; i++)
	{
		DC.AddEdge(S, i, bn[i]);
	}
	for (int i = 1; i <= n1; i++) {
		DC.AddEdge(m1 + i, T, ax[i]);
	}
	for (int i = 0; i < k; i++) {
		if (ww[i].w1<=mid) {
 			DC.AddEdge(ww[i].y, m1 + ww[i].x, ax[ww[i].x]);
		}
	}  
	if (DC.Maxflow(S, T) >= sum1) {
		return 1;
	}
	else return 0;
}
int main(void) {
	scanf("%d%d%d", &n1, &m1, &k);
	sum1 = 0;
	sum2 = 0;
	for (int i = 1; i <= n1; i++) {
		scanf("%d", &ax[i]);
		sum1 += ax[i];
	}
	for (int i = 1; i <= m1; i++) {
		scanf("%d", &bn[i]);
		sum2 += bn[i];
	}
	if (sum1 > sum2) {
		printf("-1\n");
		return 0;
	}
	for (int i = 0; i < k; i++) {
		scanf("%d%d%d", &ww[i].x, &ww[i].y, &ww[i].w1);
	}
	int left = 0;
	int right = 1000000;
	int ans = -1;
	while (left <= right) {
		int mid = (left + right) / 2;
 		if (check(mid)) {
			ans = mid;
			right = mid - 1;
		}
		else left = mid + 1;
	}
	printf("%d\n", ans);
	return 0;
}