最大权闭合图。题意我不是很理解，如果按英语原句翻译的话。我觉得输入的正值是亏损，负值是利润。这个可以参考例样，例样解雇了4、5，获得了2的利润。对于第二问如何建图，可以参考其他博客。这个需要画到纸上慢慢理解，不好解释。
对于第一问，仔细分析可以发现：
1.dfs过程不会访问到汇点，否则最大流过程没有完成。
2.如果一条路径上的员工收支平衡，那么这条路径上的所有员工都不会被访问到。
3.用标记数组标记已访问过的点之后，不需要担心最大流建图中反向边的影响。
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <memory.h>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iostream>
#include <sstream>#define ll __int64using namespace std;#define typec ll
#define inf 1000000000
#define E 65010
#define V 5005
struct Dinic {struct edge { int x, y, nxt; typec c; } bf[2 * E];int ne, head[V], cur[V], ps[V], dep[V];void init(int n) {memset(head, 0, (n+1) * sizeof(int));ne = 2;}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++;}typec flow(int n, int s, int t) {typec tr, res = 0;int i, j, k, f, r, top;while (true) {memset(dep, -1, (n+1) * 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+1) * 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;}
} D;bool vis[5005];int dfs(int x) {int ret = 0;vis[x] = 1;for (int p = D.head[x]; p != 0; p = D.bf[p].nxt) {if (!vis[D.bf[p].y] && D.bf[p].c > 0) {ret++;ret += dfs(D.bf[p].y);}}return ret;
}int main() {int n, m, x, y;ll v, sum;while (scanf("%d %d", &n, &m) != EOF) {D.init(n + 2);sum = 0;for (int i = 1; i <= n; i++) {scanf("%I64d", &v);if (v > 0) D.addedge(0, i, v), sum += v;else if (v < 0) D.addedge(i, n+1, -v);}for (int i = 0; i < m; i++) {scanf("%d %d", &x, &y);D.addedge(x, y, inf);}sum -= D.flow(n + 2, 0, n + 1);memset(vis, 0, n + 2);printf("%d %I64d\n", dfs(0), sum);}return 0;
}