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：

     

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<cstdlib>
  5 #include<cmath>
  6 #include<algorithm>
  7 #include<string>
  8 #include<map>
  9 #include<set>
 10 #include<vector>
 11 #include<queue>
 12 #include<stack>
 13 #include<ctime>
 14 #define M0(x)  memset(x, 0, sizeof(x))
 15 #define Inf 0x3fffffff
 16 #define N 700
 17 #define M 201000
 18 using namespace std;
 19 struct edge{
 20      int v, f, c, next;
 21      edge(){}
 22      edge(int _v, int _f, int _c, int _next):v(_v), f(_f), c(_c), next(_next){}
 23 };
 24 struct Costflow{
 25      int last[N], pre[N], dis[N], vis[N];
 26      int tot, S, T;
 27      edge e[M];
 28      void add(int u, int v, int f, int c){
 29           e[tot] = edge(v, f, c, last[u]); last[u] = tot++;
 30           e[tot] = edge(u, 0, -c, last[v]); last[v] = tot++; 
 31      }
 32      void init(int S, int T){
 33           this->S = S, this->T = T;
 34           memset(last, -1, sizeof(last));
 35           tot = 0;     
 36      }
 37      int spfa(){
 38           for (int i = 0; i <= T; ++i)
 39               dis[i] = Inf;
 40           memset(vis, 0, sizeof(vis));
 41           memset(pre, -1, sizeof(pre));
 42           dis[S] = 0;
 43           queue<int> q;
 44           q.push(S) , vis[S] = 1;
 45           int p, u, v;
 46           while (!q.empty()){
 47                 p = last[u = q.front()];
 48                 for ( ; p > -1; p = e[p].next){
 49                        v = e[p].v;
 50                        if (dis[u] + e[p].c < dis[v] && e[p].f > 0){
 51                              dis[v] = dis[u] + e[p].c;
 52                              pre[v] = p;
 53                              if (!vis[v]) vis[v] = 1, q.push(v);
 54                        }   
 55                 }
 56                 vis[u] = 0;
 57                 q.pop();    
 58           }
 59           return dis[T] < Inf;  
 60      }
 61      int costflow(){
 62           int cost = 0, flow = 0;
 63           while (spfa()){
 64                int f = Inf;
 65                for (int p = pre[T]; p != -1; p = pre[e[p^1].v])
 66                       f = min(f, e[p].f);
 67                flow += f;
 68                cost += f * dis[T];
 69                for (int p = pre[T]; p != -1; p = pre[e[p^1].v])
 70                       e[p].f -= f, e[p^1].f += f;     
 71           }
 72           return cost;
 73      }
 74      /***********/ 
 75 } F;
 76 int n, m;
 77 int t[70][70];
 78 
 79 void init(){
 80      for (int j = 1; j <= m; ++j)
 81         for (int i = 1; i <= n; ++i)
 82           scanf("%d", &t[i][j]);     
 83 }
 84 
 85 void solve(){
 86     int S = 0, T = n * m + m + 1;
 87     F.init(S, T);
 88     int cnt = n * m;
 89     for (int i = cnt + 1; i <= cnt + m; ++i)
 90          F.add(i, T, 1, 0);
 91     int cur = 0;
 92     for (int i = 1; i <= n; ++i)
 93        for (int j = 1; j <= m; ++j){
 94             ++cur;
 95             F.add(S, cur, 1, 0);
 96             for (int k = 1; k <= m; ++k)
 97                 F.add(cur, cnt + k, 1, j * t[i][k]);    
 98        }
 99     double ans = F.costflow();
100     printf("%.2f\n",ans / m);
101 }
102 
103 int main(){
104 //    freopen("a.in", "r", stdin);
105 //    freopen("a.out", "w", stdout);
106     while (scanf("%d%d", &n, &m) != EOF){
107           init();
108           solve();
109     }
110 }
View Code

 

posted on 2014-10-21 21:40  yzcstc  阅读(610)  评论(0)  编辑  收藏  举报