Gold Transportation

题目

百度

分析

很容易想到二分答案

然后考虑判定

条件很多，奇奇怪怪

那就上网络流吧

边权 \(\leq mid\) 两个城市连边 \(inf\)

源点与所有城市连边，边权为本城市有金矿量

城市与自己的仓库连边，边权为本城市仓库容量

仓库与汇点两边 \(inf\)

\(Code\)

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

const int N = 205 , M = 40005 , INF = 2e9;
int w[N] , c[N] , tot , n , m , W , S = 0 , T = 2 * N - 9;
int h[2 * N] , cur[2 * N] , dep[2 * N];
struct node{int x , y , z;}b[M];
struct edge{int to , nxt , w;}e[2 * M];

inline void add(int x , int y , int z){e[++tot] = edge{y , h[x] , z} , h[x] = tot;}

void build(int up)
{
	memset(h , 0 , sizeof h);
	tot = 1;
	for(register int i = 1; i <= m; i++)
	if (b[i].z <= up) add(b[i].x , b[i].y , INF) , add(b[i].x , b[i].y , 0),
				add(b[i].y , b[i].x , INF) , add(b[i].y , b[i].x , 0);
	for(register int i = 1; i <= n; i++) add(i , i + n , c[i]) , add(i + n , i , 0);
	for(register int i = 1; i <= n; i++) add(S , i , w[i]) , add(i , S , 0);
	for(register int i = 1; i <= n; i++) add(i + n , T , INF) , add(T , i + n , 0);
}

queue<int> q;
int bfs()
{
	while (!q.empty()) q.pop();
	for(register int i = S; i <= T; i++) cur[i] = h[i] , dep[i] = 0;
	dep[S] = 1 , q.push(S);
	while (!q.empty())
	{
		int now = q.front(); q.pop();
		for(register int i = h[now]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (dep[v] || e[i].w == 0) continue;
			dep[v] = dep[now] + 1 , q.push(v);
		}
	}
	return dep[T];
}

int dfs(int x , int fa , int mi)
{
	if (x == T || mi <= 0) return mi;
	int flow = 0;
	for(register int i = cur[x]; i; i = e[i].nxt)
	{
		cur[x] = i;
		int v = e[i].to;
		if (e[i].w == 0 || v == fa || dep[x] + 1 != dep[v]) continue;
		int f = dfs(v , x , min(mi , e[i].w));
		if (f <= 0) continue;
		flow += f , mi -= f , e[i].w -= f , e[i ^ 1].w += f;
		if (mi <= 0) break;
	}
	return flow;
}

int dinic()
{
	int flow = 0;
	while (bfs()) flow += dfs(S , -1 , INF);
	return flow;
}

int check(int up)
{
	build(up);
	if (dinic() >= W) return 1;
	return 0;
}

int main()
{
	scanf("%d" , &n);
	while (n != 0)
	{
		W = 0;
		for(register int i = 1; i <= n; i++) scanf("%d" , &w[i]) , W += w[i];
		for(register int i = 1; i <= n; i++) scanf("%d" , &c[i]);
		scanf("%d" , &m);
		int l = INF , r = -INF , mid , ans = INF;
		for(register int i = 1; i <= m; i++)
			scanf("%d%d%d" , &b[i].x , &b[i].y , &b[i].z),
			l = min(l , b[i].z) , r = max(r , b[i].z);
		while (l <= r)
		{
			mid = (l + r) >> 1;
			if (check(mid)) r = mid - 1 , ans = mid;
			else l = mid + 1;
		}
		if (ans == INF) printf("No Solution\n");
		else printf("%d\n" , ans);
		scanf("%d" , &n);
	}
}