Original link: https://www.shuizilong.com/house/archives/luogu-p2053-scoi2007-%E4%BF%AE%E8%BD%A6/
- https://www.luogu.com.cn/problem/P2053
It seems that dp is possible, but the cost of repairing each car is not the same for each person.
So it can be modeled as a bipartite graph.
#include <lastweapon/io>
#include <lastweapon/mincostflow>
using namespace lastweapon;
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
int m, n; RD(m, n); int s = n*(m+1), t = s+1;
mcf_graph<int, int> G(t+1);
REP(i, n) {
G. add_edge(s, i, 1, 0);
REP(j, m) {
G.add_edge(n*(j+1)+i, t, 1, 0);
int c; RD(c); REP(k, n) G.add_edge(i, n*(j+1)+k, 1, (k+1)*c);
}
}
printf("%.2f\n", (DB)G.flow(s, t).se / n);
}
#include <lastweapon/mincostflow2>
using namespace lastweapon;
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
int m, n; RD(m, n); int s = n*(m+1), t = s+1;
mcf_graph<int, int, 2047, 105536, 2047> G;
Gs = s; Gt = t; Gn = t+1;
REP(i, n) {
G. add_edge(s, i, 1, 0);
REP(j, m) {
G.add_edge(n*(j+1)+i, t, 1, 0);
int c; RD(c); REP(k, n) G.add_edge(i, n*(j+1)+k, 1, (k+1)*c);
}
}
printf("%.2f\n", (DB)G.Run().se / n);
}
This article is reproduced from: https://www.shuizilong.com/house/archives/luogu-p2053-scoi2007-%E4%BF%AE%E8%BD%A6/
This site is only for collection, and the copyright belongs to the original author.