【BZOJ】1070: [SCOI2007]修车（费用流+特殊的技巧）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }

const int N=2005, oo=0x7f7f7f7f;
int ihead[N], cnt=1, q[N], n, p[N], d[N], vis[N];
struct dat { int next, to, cap, from, w; }e[N*N];
void add(int u, int v, int c, int w) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u; e[cnt].cap=c; e[cnt].w=w;
	e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v; e[cnt].cap=0; e[cnt].w=-w;
}
bool spfa(int s, int t) {
	memset(vis, 0, sizeof(int)*(t+1));
	memset(d, 0x7f, sizeof(int)*(t+1));
	d[s]=0; int front=0, tail=0;
	q[tail++]=s;
	while(front!=tail) {
		int u=q[front++], v; if(front==N) front=0; vis[u]=0;
		rdm(u, i) if(e[i].cap) {
			v=e[i].to;
			if(d[v]>d[u]+e[i].w) {
				d[v]=d[u]+e[i].w;
				p[v]=i;
				if(!vis[v]) {
					vis[v]=1;
					if(d[v]<d[q[front]]) {
						--front; if(front<0) front+=N;
						q[front]=v;
					}
					else {
						q[tail++]=v; if(tail==N) tail=0;
					}
				}
			}
		}
	}
	return d[t]!=oo;
}
int mcf(int s, int t) {
	int ret=0, f, u;
	while(spfa(s, t)) {
		f=oo;
		for(u=t; u!=s; u=e[p[u]].from) f=min(f, e[p[u]].cap);
		for(u=t; u!=s; u=e[p[u]].from) e[p[u]].cap-=f, e[p[u]^1].cap+=f;
		ret+=f*d[t];
	}
	return ret;
}
int id, a[65][65];
int main() {
	read(n); int m=getint();
	id=m;
	for1(i, 1, m) for1(j, 1, n) read(a[j][i]);
	for1(i, 1, n) for1(k, 1, m) { ++id; for1(j, 1, m) add(id, j, 1, k*a[i][j]); }
	int s=id+1, t=s+1;
	for1(i, 1, m) add(i, t, 1, 0);
	for1(i, m+1, id) add(s, i, 1, 0);
	printf("%.2f\n", (double)mcf(s, t)/m);
	return 0;
}