ACM/ICPC 之 网络流-拆点构图(POJ2391)

　　需要直接到达，因此源点经过三条边后必须要达到汇点，但为了保证网络流的正确性（路径可反悔），因此不可限制层次网络的最高层次为3，最好的方法既是让所有点拆分成两个点，一个点从汇点进入，一个点通向汇点，任意两点的路径则标注为最短路径。

 

//Dinic算法-拆点构图
//Time:625Ms    Memory:2108K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
#define MAX 405
#define INF 0x3f3f3f3f
#define LL long long
int n, m;
int s, t;
int cow[MAX], hide[MAX];
LL d[MAX][MAX];
int res[MAX][MAX];
int lev[MAX];
void build_map(LL limit)	//拆点构图
{
	memset(res, 0, sizeof(res));
	for (int i = 1; i <= n; i++)
	{
		res[s][i] = cow[i];
		res[i + n][t] = hide[i];
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			if (d[i][j] <= limit) res[i][j + n] = INF;
}
bool bfs()	//构建层次网络
{
	memset(lev, -1, sizeof(lev));
	queue<int> q;
	q.push(s);  lev[s] = 0;
	while (!q.empty() && lev[t] == -1) {
		int cur = q.front();    q.pop();
		for (int i = 1; i <= t; i++)
		{
			if (lev[i] == -1 && res[cur][i])
			{
				lev[i] = lev[cur] + 1;
				q.push(i);
			}
		}
	}
	return lev[t] != -1;
}
int dfs(int x, int sum)	//增广并更新
{
	if (x == t || sum == 0) return sum;
	int src = sum;
	for (int i = 1; i <= t; i++)
	{
		if (lev[i] == lev[x] + 1 && res[x][i])
		{
			int tmp = dfs(i, min(sum, res[x][i]));
			res[x][i] -= tmp;
			res[i][x] += tmp;
			sum -= tmp;
		}
	}
	return src - sum;
}
int Dinic()
{
	int maxFlow = 0;
	while (bfs())
		maxFlow += dfs(0, INF);
	return maxFlow;
}
int main()
{
	//freopen("in.txt", "r", stdin);
	memset(d, INF, sizeof(d));
	scanf("%d%d", &n, &m);
	s = 0; t = 2 * n + 1;
	int cows = 0;   //总牛数
	for (int i = 1; i <= n; i++)
	{
		scanf("%d%d", &cow[i], &hide[i]);
		cows += cow[i];
	}
	for (int i = 1; i <= m; i++)
	{
		int u, v;
		LL w;
		scanf("%d%d%lld", &u, &v, &w);
		d[u][v] = d[v][u] = min(w, d[u][v]);
	}
	//Floyd
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
		{
			d[0][i] = d[i][t] = 0;
			if (d[i][k] != d[0][0]) {
				for (int j = 1; j <= n; j++)
				{
					if (i != j)  d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
					else d[i][j] = 0;
				}
			}
		}
	LL l = 0, r = 200LL * 1000000000;
	int last;
	while (l < r)
	{
		LL mid = (l + r) / 2;
		build_map(mid);
		last = Dinic();
		last == cows ? r = mid : l = mid + 1;
	}
	if (r != 200LL * 1000000000)
		printf("%lld\n", r);
	else printf("-1\n");
	return 0;
}