bzoj1070

题意：给定m个工人，n个车，并给出现在每个工人洗每辆车的时间。。如果这n辆车同时到达，问平均最小的等待时间（到达到洗完）是多少？

思路：

      很妙的构图！

      先拆点，把一个工人拆成n个。那么第i个工人倒数第j次时洗第k辆车贡献的总等待时间为j*t[i][k]。

     那么接下来就是一个不难想的费用流了。

code：

     

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<ctime>
#define M0(x)  memset(x, 0, sizeof(x))
#define Inf 0x3fffffff
#define N 700
#define M 201000
using namespace std;
struct edge{
     int v, f, c, next;
     edge(){}
     edge(int _v, int _f, int _c, int _next):v(_v), f(_f), c(_c), next(_next){}
};
struct Costflow{
     int last[N], pre[N], dis[N], vis[N];
     int tot, S, T;
     edge e[M];
     void add(int u, int v, int f, int c){
          e[tot] = edge(v, f, c, last[u]); last[u] = tot++;
          e[tot] = edge(u, 0, -c, last[v]); last[v] = tot++; 
     }
     void init(int S, int T){
          this->S = S, this->T = T;
          memset(last, -1, sizeof(last));
          tot = 0;     
     }
     int spfa(){
          for (int i = 0; i <= T; ++i)
              dis[i] = Inf;
          memset(vis, 0, sizeof(vis));
          memset(pre, -1, sizeof(pre));
          dis[S] = 0;
          queue<int> q;
          q.push(S) , vis[S] = 1;
          int p, u, v;
          while (!q.empty()){
                p = last[u = q.front()];
                for ( ; p > -1; p = e[p].next){
                       v = e[p].v;
                       if (dis[u] + e[p].c < dis[v] && e[p].f > 0){
                             dis[v] = dis[u] + e[p].c;
                             pre[v] = p;
                             if (!vis[v]) vis[v] = 1, q.push(v);
                       }   
                }
                vis[u] = 0;
                q.pop();    
          }
          return dis[T] < Inf;  
     }
     int costflow(){
          int cost = 0, flow = 0;
          while (spfa()){
               int f = Inf;
               for (int p = pre[T]; p != -1; p = pre[e[p^1].v])
                      f = min(f, e[p].f);
               flow += f;
               cost += f * dis[T];
               for (int p = pre[T]; p != -1; p = pre[e[p^1].v])
                      e[p].f -= f, e[p^1].f += f;     
          }
          return cost;
     }
     /***********/ 
} F;
int n, m;
int t[70][70];

void init(){
     for (int j = 1; j <= m; ++j)
        for (int i = 1; i <= n; ++i)
          scanf("%d", &t[i][j]);     
}

void solve(){
    int S = 0, T = n * m + m + 1;
    F.init(S, T);
    int cnt = n * m;
    for (int i = cnt + 1; i <= cnt + m; ++i)
         F.add(i, T, 1, 0);
    int cur = 0;
    for (int i = 1; i <= n; ++i)
       for (int j = 1; j <= m; ++j){
            ++cur;
            F.add(S, cur, 1, 0);
            for (int k = 1; k <= m; ++k)
                F.add(cur, cnt + k, 1, j * t[i][k]);    
       }
    double ans = F.costflow();
    printf("%.2f
",ans / m);
}

int main(){
//    freopen("a.in", "r", stdin);
//    freopen("a.out", "w", stdout);
    while (scanf("%d%d", &n, &m) != EOF){
          init();
          solve();
    }
}