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传送门:http://poj.org/problem?id=2987
题目大意:
公司裁员,每裁一个员工有相应的收益(正负都有可能),当一个员工被裁,他的下属也全会被裁。求最大的收益以及此时被裁的最少人数。
思路:
首先增加一个源点S和汇点T,S指向所有收益>0的点,所有收益<0的点指向T,再将每个员工/领导关系以领导指向员工建图,这样来构建一个二分图,求它的最大权闭合图。
由于最大权闭合图=最大权之和(所有正的权相加)-最小割,而根据最大流最小割定理可以知道最大流就等于最小割,因此可以转变为求最大流。
统计人数只需要从S搜索没有满流的那一块的点数即可。
另外需要注意开long long。。
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AC代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs id*2+1
#define Mod(a,b) a<b?a:a%b+b
using namespace std;
const ll M = 1e9 + 7;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 5010;
const double e = 10e-6;
struct edge
{
int to;
ll cap;
int rev;
};
vector<edge>G[N];
int level[N], iter[N];
void addEdge(int from, int to, ll cap)
{
G[from].push_back(edge{ to,cap,(int)G[to].size() });
G[to].push_back(edge{ from,0,(int)G[from].size() - 1 });
}
void bfs(int s)
{
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0;
que.push(s);
while (!que.empty()) {
int v = que.front(); que.pop();
int len = G[v].size();
for (int i = 0; i < len; i++) {
edge &e = G[v][i];
if (e.cap > 0 && level[e.to] < 0) {
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}
ll dfs(int v, int t, ll f)
{
if (v == t)
return f;
int len = G[v].size();
for (int &i = iter[v]; i < len; i++) {
edge &e = G[v][i];
if (e.cap > 0 && level[v] < level[e.to]) {
ll d = dfs(e.to, t, min(f, e.cap));
if (d > 0) {
e.cap -= d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
ll maxFlow(int s, int t)
{
ll flow = 0;
while (1) {
bfs(s);
if (level[t] < 0)
return flow;
memset(iter, 0, sizeof(iter));
ll f;
while ((f = dfs(s, t, INF)) > 0)
flow += f;
}
}
int n, m, num;
bool vis[N];
void cal(int u)
{
vis[u] = true;
num++;
int len = G[u].size();
for (int i = 0; i < len; i++) {
edge e = G[u][i];
if (e.cap > 0 && !vis[e.to])
cal(e.to);
}
}
int main() {
while (~scanf("%d%d", &n, &m)) {
for (int i = 0; i <= n + 1; i++)
G[i].clear();
ll xx, ans = 0;
for (int i = 1; i <= n; i++) {
scanf("%lld", &xx);
if (xx > 0) {
addEdge(0, i, xx);
ans += xx;
}
else if (xx < 0)
addEdge(i, n + 1, -xx);
}
int x, y;
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x, &y);
addEdge(x, y, INF);
}
ans -= maxFlow(0, n + 1);
memset(vis, false, sizeof(vis));
num = 0;
cal(0);
printf("%d %lld\n", num - 1, ans);
}
return 0;
}