poj3469

网络流的最小割问题，看到分配成两部分的题，就要想到网络流最小割，割值是一部分指向另一部分的边的初始容量的总和，不加上另一部分指向本部分的边的容量。而对于最小割来说，最大流流量就是最小割的容量。而划分方法，就是将最大流后残余网络中s可以走到的点划为s集合，其余点划为t集合，注意，不是可以走到t的点，因为有可能有一条线上的多条边都流满的情况，这种情况下有些点就无法走到t，而且s也走不到它。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define N 20005
#define M 500005
#define typec int
const typec inf = 0x3f3f3f3f;

struct edge
{
	int x, y, nxt;
	typec c;
} bf[M];
int ne, head[N], cur[N], ps[N], dep[N];

void addedge(int x, int y, typec c)
{
	bf[ne].x = x;
	bf[ne].y = y;
	bf[ne].c = c;
	bf[ne].nxt = head[x];
	head[x] = ne++;
	bf[ne].x = y;
	bf[ne].y = x;
	bf[ne].c = 0;
	bf[ne].nxt = head[y];
	head[y] = ne++;
}
void addedge1(int x, int y, typec c)
{
	bf[ne].x = x;
	bf[ne].y = y;
	bf[ne].c = c;
	bf[ne].nxt = head[x];
	head[x] = ne++;
	bf[ne].x = y;
	bf[ne].y = x;
	bf[ne].c = c;
	bf[ne].nxt = head[y];
	head[y] = ne++;
}

typec flow(int n, int s, int t)
{
	typec tr, res = 0;
	int i, j, k, f, r, top;
	while (1)
	{
		memset(dep, -1, n * sizeof(int));
		for (f = dep[ps[0] = s] = 0, r = 1; f != r;)
		{
			for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
			{
				if (bf[j].c && -1 == dep[k = bf[j].y])
				{
					dep[k] = dep[i] + 1;
					ps[r++] = k;
					if (k == t)
					{
						f = r;
						break;
					}
				}
			}
		}
		if (-1 == dep[t])
			break;

		memcpy(cur, head, n * sizeof(int));
		for (i = s, top = 0;;)
		{
			if (i == t)
			{
				for (k = 0, tr = inf; k < top; ++k)
					if (bf[ps[k]].c < tr)
						tr = bf[ps[f = k]].c;
				for (k = 0; k < top; ++k)
					bf[ps[k]].c -= tr, bf[ps[k] ^ 1].c += tr;
				res += tr;
				i = bf[ps[top = f]].x;
			}
			for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
				if (bf[j].c && dep[i] + 1 == dep[bf[j].y])
					break;
			if (cur[i])
			{
				ps[top++] = cur[i];
				i = bf[cur[i]].y;
			}
			else
			{
				if (0 == top)
					break;
				dep[i] = -1;
				i = bf[ps[--top]].x;
			}
		}
	}
	return res;
}
int main()
{
	//freopen("D:\\t.txt", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	ne = 2;
	memset(head, 0, sizeof(head));
	n += 2;
	int s = 0;
	int t = 1;
	for (int i = 2; i < n; i++)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		addedge(s, i, a);
		addedge(i, t, b);
	}
	for (int i = 0; i < m; i++)
	{
		int x, y, c;
		scanf("%d%d%d", &x, &y, &c);
		addedge1(x + 1, y + 1, c);
	}
	printf("%d\n", flow(n, s, t));
	return 0;
}