[SCOI2007]修车 费用流 BZOJ 1070
题目描述
同一时刻有N位车主带着他们的爱车来到了汽车维修中心。维修中心共有M位技术人员，不同的技术人员对不同的车进行维修所用的时间是不同的。现在需要安排这M位技术人员所维修的车及顺序，使得顾客平均等待的时间最小。
说明：顾客的等待时间是指从他把车送至维修中心到维修完毕所用的时间。
输入输出格式
输入格式：第一行有两个数M,N，表示技术人员数与顾客数。
接下来n行，每行m个整数。第i+1行第j个数表示第j位技术人员维修第i辆车需要用的时间T。
输出格式：最小平均等待时间，答案精确到小数点后2位。
输入输出样例
说明
(2<=M<=9,1<=N<=60), (1<=T<=1000)
假设对于某一个技术工人来说，维修序列为：
w1,w2,w3...,wn;
那么等待的时间：
T=n*w1+(n-1)*w2+...+wn;
可见费用（消耗的时间）与位置有关；
那么我们将技术工拆点，拆成 n 个；
然后建边的时候分别设立不同的费用，最后跑一下最小费用最大流；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 20005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
bool vis[maxn];
int n, m, s, t;
int x, y, f, z;
int dis[maxn], pre[maxn], last[maxn], flow[maxn];
int maxflow, mincost;
struct node {
int to, nxt, flow, cost;
}edge[maxn<<2];
int head[maxn], cnt;
queue<int>q;
void addedge(int from, int to, int flow, int cost) {
edge[++cnt].to = to; edge[cnt].flow = flow; edge[cnt].cost = cost;
edge[cnt].nxt = head[from]; head[from] = cnt;
}
bool spfa(int s, int t) {
memset(dis, 0x7f, sizeof(dis)); memset(flow, 0x7f, sizeof(flow));
ms(vis);
q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
while (!q.empty()) {
int now = q.front(); q.pop(); vis[now] = 0;
for (int i = head[now]; i != -1; i = edge[i].nxt) {
if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].cost) {
int v = edge[i].to;
dis[v] = dis[now] + edge[i].cost;
pre[v] = now; last[v] = i;
flow[v] = min(flow[now], edge[i].flow);
if (!vis[v]) {
vis[v] = 1; q.push(v);
}
}
}
}
return pre[t] != -1;
}
void mincost_maxflow() {
while (spfa(s, t)) {
int now = t;
maxflow += flow[t]; mincost+=flow[t] * dis[t];
while (now != s) {
edge[last[now]].flow -= flow[t];
edge[last[now] ^ 1].flow += flow[t];
now = pre[now];
}
}
}
int main() {
//ios::sync_with_stdio(0);
memset(head, -1, sizeof(head)); cnt = 1;
rdint(m); rdint(n);
s = 1000; t = s + 1;
for (int i = 1; i <= n; i++)addedge(s, i, 1, 0), addedge(i, s, 0, 0);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
int tmp; rdint(tmp);
for (int k = 1; k <= n; k++) {
addedge(i, j*n + k, 1, tmp*k); addedge(j*n + k, i, 0, -tmp * k);
}
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++)addedge(i*n + j, t, 1, 0), addedge(t, i*n + j, 0, 0);
}
mincost_maxflow();
printf("%.2lf\n", 1.0*mincost / n);
return 0;
}
EPFL - Fighting
