pku 2987

题目其实是求最大闭合图，将原问题的图转化为最小割模型。

在原图点集合基础上添加源st和汇ed，将原图每条边(u, v)容量设为oo，增加源点st到原图每个正权点v的有向边，容量为node[v]，增加原图每个负权点到汇点ed的有向边，容量为-node[v]，然后求st到ed的最大流，在残留网络中与源点相连的点为被选择的点，所有正权点权值之和减去最大流为题中所求最大获益，题目权值的绝对值小于10000000，得用long long，呵呵，在这点上WA了一次。

代码：

#include <stdio.h>
#include <string.h>
#define oo 0x7fffffff
#define MAXM 100002
#define MAXN 5005
typedef long long int64;
typedef struct
{
	int u; int v; int cap; int next;
}edge_t;
edge_t e[MAXM];
int first[MAXN];
int node[MAXN];
int dist[MAXN];
int now[MAXN];
int pre[MAXN];
int cnt[MAXN];
int cur[MAXN];
int tot, st, ed, n2, n;
int64 sum;
void add(int u, int v, int cap)
{
	e[++tot].u = u; e[tot].v = v; e[tot].cap = cap; e[tot].next = first[u]; first[u] = tot;
	e[++tot].u = v; e[tot].v = u; e[tot].cap = 0;   e[tot].next = first[v]; first[v] = tot;
}
void init()
{
	int i, j, m;
	scanf("%d %d", &n, &m);
	memset(first, -1, sizeof(first));
	sum = 0; tot = -1; st = 0; ed = n + 1;
	for(i = 1;i <= n; i++)
	{
		scanf("%d", &node[i]);
		if(node[i] > 0) { add(st, i, node[i]); sum += node[i];}
		else add(i, ed, -node[i]);
	}
	while(m--)
	{
		scanf("%d %d", &i, &j);
		add(i, j, oo);
	}
	n2 = n; n += 2;
}
int64 sap()
{
	int64 tot_flow;
	int now_flow, found, min;
	int i, j, t;
	memset(dist, 0, sizeof(dist));
	memset(now, -1, sizeof(now));
	memset(cnt, 0, sizeof(cnt));
	i = st;
	tot_flow = 0; now_flow = oo;
	cnt[0] = n;
	while(dist[st] < n)
	{
		cur[i] = now_flow; found = 0;
		if(now[i] == -1) t = first[i];
		else t = now[i];
		while(t != -1)
		{
			j = e[t].v;
			if(e[t].cap > 0 && dist[j] + 1 == dist[i])
			{
				found = 1; now[i] = t;
				if(e[t].cap < now_flow) now_flow = e[t].cap;
				pre[j] = t; i = j;
				if(i == ed)
				{
					tot_flow += now_flow;
					while(i != st)
					{
						e[pre[i]].cap -= now_flow;
						e[pre[i]^1].cap += now_flow;
						i = e[pre[i]].u;
					}
					now_flow = oo;
				}
				break;
			}
			t = e[t].next;
		}
		if(found) continue;
		if(--cnt[dist[i]] == 0) break;
		min = n - 1;
		t = first[i];
		while(t != -1)
		{
			if(e[t].cap > 0 && dist[e[t].v] < min)
			{
				min = dist[e[t].v];
				now[i] = t;
			}
			t = e[t].next;
		}
		dist[i] = min + 1;
		cnt[dist[i]]++;
		if(i != st)
		{
			i = e[pre[i]].u;
			now_flow = cur[i];
		}
	}
	return tot_flow;
}
void dfs(int u)
{
	int t, v;
	cnt[u] = 1;
	t = first[u];
	while(t != -1)
	{
		v = e[t].v;
		if(e[t].cap > 0 && cnt[v] == 0) dfs(v);
		t = e[t].next;
	}
}
void solve()
{
	int64 res;
	int ans, i;
	res = sap();
	memset(cnt, 0, sizeof(cnt));
	dfs(st); ans = 0;
	for(i = 1;i <= n2; i++) ans += cnt[i];
	printf("%d %lld\n", ans, sum - res);
}
int main()
{
	init();
	solve();
	return 0;
}