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 }
