标签：pop freopen http ios class out insert space targe
题目链接：New Equipments
题意：有n个工人，m台机器，每个工人有三个属性$a_i,b_i,c_i$，现在要把工人安排到机器上工作，一个工人只能安排到一个机器，一个机器上也只能安排一个工人，把第$i$个工人安排到第$j$个机器上的代价为$a_i \times j^2 + b_i \times j + c_i$，分别求安排[1 $\cdots$ n]个工人的最小代价
思路：由于每个工人的代价为二次函数，我们可以利用二元一次函数的性质求出代价的最小值，然后向两边扩展，总共找出n个代价最小的机器，将这个工人和这n台机器之间连一条容量为1，费用为$a_i \times j^2 + b_i \times j + c_i$的边，对每个工人都这么操作，然后跑费用流即可，由于要分别求出安排[1 $\cdots$ n]个工人的最小代价，所以在残余网络上跑n次spfa即可
参考：https://blog.csdn.net/qq_43906000/article/details/107545485
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <cmath> using namespace std; typedef long long ll; const int N = 5010; const int M = 10010; const int NN = 55; const ll INF = 1000000000000000010; struct Edge { int to, nex; ll w, c; }; Edge edge[2 * M]; int T, n, m, mx, p[NN][NN]; ll a[NN], b[NN], c[NN]; vector<int> alls; int cnt, s, t, pre[N], lst[N], head[N], inq[N]; ll mc, dis[N], mf, f[N]; queue<int> q; void init() { mc = mf = 0; alls.clear(); cnt = -1; memset(head, -1, sizeof(head)); } void add_edge(int u, int v, ll w, ll c) { edge[++cnt].to = v; edge[cnt].w = w; edge[cnt].c = c; edge[cnt].nex = head[u]; head[u] = cnt; } int spfa(int s, int t) { for (int i = 0; i < N; i++) { dis[i] = f[i] = INF; inq[i] = 0; } while (!q.empty()) q.pop(); q.push(s); inq[s] = 1, dis[s] = 0, pre[t] = -1; while (!q.empty()) { int u = q.front(); q.pop(); inq[u] = 0; for (int i = head[u]; -1 != i; i = edge[i].nex) { int v = edge[i].to; ll c = edge[i].c, w = edge[i].w; if (0 == w || dis[v] <= dis[u] + c) continue; dis[v] = dis[u] + c; pre[v] = u; lst[v] = i; f[v] = min(f[u], w); if (1 == inq[v]) continue; inq[v] = 1; q.push(v); } } if (-1 == pre[t]) return 0; return 1; } void insert(int u, int v, ll w, ll c) { add_edge(u, v, w, c); add_edge(v, u, 0, -c); } ll fx(int i, ll x) { return x * x * a[i] + x * b[i] + c[i]; } int fid(int x) { return lower_bound(alls.begin(), alls.end(), x) - alls.begin() + 1; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { init(); scanf("%d%d", &n, &mx); for (int i = 1; i <= n; i++) { scanf("%lld%lld%lld", &a[i], &b[i], &c[i]); int res = 0, c = 1; if (b[i] >= 0) res = 1; else { int lm = floor(-1.0 * b[i] / (2 * a[i])); int rm = ceil(-1.0 * b[i] / (2 * a[i])); if (lm >= 1) { if (fx(i, lm) < fx(i, rm)) res = lm; else res = rm; } else res = rm; } p[i][c] = res; int l = res - 1, r = res + 1; while (c < n) { if (fx(i, l) < fx(i, r)) { if (l >= 1 && l <= mx) p[i][++c] = l--; else p[i][++c] = r++; } else { if (r >= 1 && r <= mx) p[i][++c] = r++; else p[i][++c] = l--; } } } for (int i = 1; i <= n; i++) for (int k = 1; k <= n; k++) alls.push_back(p[i][k]); sort(alls.begin(), alls.end()); alls.erase(unique(alls.begin(), alls.end()), alls.end()); m = alls.size(), s = n + m + 1, t = s + 1; for (int i = 1; i <= n; i++) { insert(s, i, 1, 0); for (int k = 1; k <= n; k++) { insert(i, n + fid(p[i][k]), 1, fx(i, p[i][k])); } } for (int i = 1; i <= m; i++) insert(n + i, t, 1, 0); for (int i = 1; i <= n; i++) { if (!spfa(s, t)) break; mf += f[t]; mc += f[t] * dis[t]; int now = t; while (now != s) { edge[lst[now]].w -= f[t]; edge[lst[now] ^ 1].w += f[t]; now = pre[now]; } printf("%lld", mc); printf(i == n ? "\n" : " "); } } return 0; }
求二元一次函数的最小值也可以用三分来实现，代码如下
int l = 1, r = mx, c = 1; while (l < r) { int lmid = floor(1.0 * (l + r) / 2); int rmid = floor(1.0 * (lmid + r) / 2); if (fx(i, lmid) <= fx(i, rmid)) r = rmid; else l = lmid; } p[i][c] = l; r = l + 1, l = l - 1;
2020 Multi-University Training Contest 2 - E. New Equipments(网络流)
标签：pop freopen http ios class out insert space targe
原文地址：https://www.cnblogs.com/zzzzzzy/p/13371030.html
