题意:n(0 < n ≤ 5000)个人,m(0 ≤ m ≤ 60000)个上下级关系,炒一个人可以获得收益或者损失bi (|bi| ≤ 10 ^ 7, 1 ≤ i ≤ n),炒一个人会把他的所有下级一起炒掉,问怎样炒人使收益最大,输出最大收益和最少炒人的数量。
明显的最大权闭包图的条件。求一次最小割即可。对于要求最少的炒人数,也就是说我们选哪个闭合图会使得总收益最大。最小割中的S集合=s交V1,该V1即为所求。
因此我们求完最小割以后用dfs从s开始遍历,能走到的点都为V1中的点。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#define MAX 5100
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
int ss, tt;
int kase;
int sum;
int n, m;
int cont;
int vis[MAX];
int head[MAX];
int divv[MAX];
int cur[MAX];
int cnt;
ll postive;
struct edge {
int from, to, next;
ll w;
}e[500000];
void add(int u, int v, int w) {
e[cont].from = u;
e[cont].to = v;
e[cont].w = w;
e[cont].next = head[u];
head[u] = cont++;
}
int makediv() {
memset(divv, 0, sizeof(divv));
divv[ss] = 1;
queue<int> Q;
Q.push(ss);
while (!Q.empty()) {
int u = Q.front();
if (u == tt)
return 1;
Q.pop();
for (int i = head[u]; i != -1; i = e[i].next) {
ll w = e[i].w;
int v = e[i].to;
if (divv[v] == 0 && w) {
divv[v] = divv[u] + 1;
Q.push(v);
}
}
}
return 0;
}
ll DFS(int u, ll maxflow, int tt) {
if (u == tt)
return maxflow;
int ret = 0;
for (int &i = cur[u]; i != -1; i = e[i].next) {
int v = e[i].to;
ll w = e[i].w;
if (divv[v] == divv[u] + 1 && w) {
int f = DFS(v, min(maxflow - ret, w), tt);
e[i].w -= f;
e[i ^ 1].w += f;
ret += f;
if (ret == maxflow)
return ret;
}
}
return ret;
}
void dfs(int s) {
vis[s] = 1;
for (int i = head[s]; i != -1; i = e[i].next) {
int v = e[i].to;
ll w = e[i].w;
if (!vis[v] && w) {
dfs(v);
cnt++;
}
}
}
void Dinic() {
ll ans = 0;
while (makediv() == 1) {
memcpy(cur, head, sizeof(head));
ans += DFS(ss, INF, tt);
}
dfs(ss);
printf("%d %I64d\n", cnt, postive - ans);
}
int main(void) {
while (~scanf("%d%d", &n, &m)) {
memset(vis, 0, sizeof(vis));
cnt = 0;
postive = 0;
cont = 0;
memset(head, -1, sizeof(head));
ss = 0;
tt = n + 1;
ll w, u, v;
for (int i = 1; i <= n; i++) {
scanf("%I64d", &w);
if (w > 0) {
add(ss, i, w);
add(i, ss, 0);
postive += w;
}
if (w < 0) {
add(i, tt, -w);
add(tt, i, 0);
}
}
for (int i = 1; i <= m; i++) {
scanf("%I64d%I64d", &u, &v);
add(u, v, INF);
add(v, u, 0);
}
Dinic();
}
return 0;
}