题目大意
给出 n n n 个工人和 m m m 件装备，装备的编号为 1 , 2 , 3... m 1, 2, 3 ... m 1,2,3...m。
对于工人 i i i ，他有三个参数 a i , b i , c i a_i, b_i, c_i ai,bi,ci，当为这个工人装备了第 j j j 个装备时，需要花费 a i ∗ j 2 + b i ∗ j + c i a_i * j ^ 2+ b_i * j + c_i ai∗j2+bi∗j+ci 的费用。
当为 k k k 个工人装备上装备时，最小花费是多少。
对所有的 k k k 的情况均需要输出
分析
费用流
将每个员工与源点链接，取每个员工的二次函数曲线的最小值附近的 n n n 个点，与员工相连，所有的在二次函数曲线上的点与汇点相连，员工连接到二次函数的线需要设定费用，其他线费用均为 0 0 0，所有线的流量均为 1 1 1
输出每次 spfa 过程中得到的费用即可
AC code
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn = 3000;
const ll INF = 0x3f3f3f3f3f3f3f3f;
bool flag;
struct Edge {
ll from, to, cap, flow, cost;
Edge(ll u, ll v, ll c, ll f, ll cc)
: from(u), to(v), cap(c), flow(f), cost(cc) {
}
};
struct MCMF {
ll n, m;
vector<Edge> edges;
vector<ll> G[maxn];
ll inq[maxn]; //是否在队列中
ll d[maxn]; //bellmanford
ll p[maxn]; //上一条弧
ll a[maxn]; //可改进量
void init(ll nn) {
this->n = nn;
for (ll i = 0; i <= n; ++i) G[i].clear();
edges.clear();
}
void addEdge(ll from, ll to, ll cap, ll cost) {
edges.emplace_back(Edge(from, to, cap, 0, cost));
edges.emplace_back(Edge(to, from, 0, 0, -cost));
m = (ll) (edges.size());
G[from].emplace_back(m - 2);
G[to].emplace_back(m - 1);
}
bool spfa(ll s, ll t, ll &flow, ll &cost) {
for (ll i = 1; i <= n; ++i) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0;
inq[s] = 1;
p[s] = 0;
queue<ll> q;
a[s] = INF;
q.push(s);
while (!q.empty()) {
ll u = q.front();
q.pop();
inq[u] = 0;
for (ll i = 0; i < (ll) (G[u].size()); ++i) {
Edge &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if (d[t] == INF) return false;
flow += a[t];
cost += (ll) d[t] * (ll) a[t];
cout << (flag ? " " : "") << cost;
flag = true;
for (ll u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u] ^ 1].flow -= a[t];
}
return true;
}
ll MincostMaxflow(ll s, ll t, ll &cost) {
ll flow = 0;
cost = 0;
while (spfa(s, t, flow, cost));
return flow;
}
} mcmf;
struct Node {
ll a, b, c;
ll l, r;
ll cal(ll x) {
return a * x * x + b * x + c;
}
void make(ll n, ll m) {
ll mid = -b / 2 / a;
l = mid - n / 2 - 1;
r = mid + n / 2 + 1;
l = max(1ll, l);
r = min(m, r);
if (r == m) {
l = r - n - 2;
} else if (l == 1) {
r = l + n + 2;
}
}
} node[60];
void solve() {
ll T;
cin >> T;
for (ll ts = 0; ts < T; ++ts) {
flag = false;
unordered_map<ll, ll> trans;
ll n, m;
cin >> n >> m;
ll ind = 51;
for (ll i = 1; i <= n; ++i) {
cin >> node[i].a >> node[i].b >> node[i].c;
node[i].make(n, m);
ll l = node[i].l;
ll r = node[i].r;
assert(r - l < 60);
for (ll j = l; j <= r; ++j) {
if (trans.count(j)) continue;
trans.insert({
j, ind++});
}
}
ll source = ind + 10, target = ind + 11;
mcmf.init(target + 10);
#ifdef ACM_LOCAL
cerr << target + 10 << endl;
#endif
for (auto &item : trans)
mcmf.addEdge(item.second, target, 1, 0);
for (ll i = 1; i <= n; ++i) {
mcmf.addEdge(source, i, 1, 0);
for (ll j = node[i].l; j <= node[i].r; ++j) {
mcmf.addEdge(i, trans[j], 1, node[i].cal(j));
}
}
ll cost;
mcmf.MincostMaxflow(source, target, cost);
cout << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
signed test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
if (acm_local_for_debug == '$') exit(0);
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}