POJ 3686 The Windy's（思維+費用流好題）
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 5362 | Accepted: 2249 |
Description
The Windy‘s is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receives N orders for toys. The manager knows that every order will take different amount of hours in different workshops. More precisely, the i
The manager wants to minimize the average of the finishing time of the N
Input
The first line of input is the number of test case. The first line of each test case contains two integers, N and M (1 ≤ N,M ≤ 50).
The next N lines each contain M integers, describing the matrix Zij (1 ≤ Zij ≤ 100,000) There is a blank line before each test case.
Output
For each test case output the answer on a single line. The result should be rounded to six decimal places.
Sample Input
3 3 4 100 100 100 1 99 99 99 1 98 98 98 1 3 4 1 100 100 100 99 1 99 99 98 98 1 98 3 4 1 100 100 100 1 99 99 99 98 1 98 98
Sample Output
2.000000 1.000000 1.333333
題目連接：POJ 3686
刷白書的網絡流問題看到的，跟某一道導彈發射的問題很像，但是這題不同工廠對不同的物品都有不同的加工時間，按加工次序拆點是很容易想到， 但這樣一想有一個問題，比如我把物品連到工場2的第二次加工時間點，那我這條邊流量是1，費用是多少根本不知道，因為我不知道工廠2第一次加工的那條邊費用是多少，即不知道當前物品加工時等待了多久，因此需要換一種思路。
比如有三個物品a,b,c都在同一個工廠加工，那麽時間可以這麽寫：ta+ta+tb+ta+tb+tc=3ta+2tb+1tc，顯然最先加工的是a，其次是b，最後是c，可以發現物品前面的系數的最大值就是在工廠加工的總物品數，那麽這題實際上就是一個系數分配問題，那麽建圖就好辦了， 費用就是加工次序*加工時間，由於流量是1，確實解決了同一時間只能加工一個物品的問題，但可能又會想，那我的某一個物品萬一在工廠1的第k次序加工，但是實際上工廠1前k-1次機會都沒有用過，這豈不是非常浪費嗎？實際上由於用的是SPFA最短路增廣，求解過程就是求出了總時間最優的情況，不會出現前面空著的加工次序問題，如果有空著的加工次序，那麽SPFA一定會找到並把這條邊松弛算進網絡流中
代碼：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 55;
const int MAXV = N + N * N;
const int MAXE = N + N * N + N * N * N;
struct edge
{
int to, nxt, cap, cost;
edge() {}
edge(int _to, int _nxt, int _cap, int _cost): to(_to), nxt(_nxt), cap(_cap), cost(_cost) {}
};
edge E[MAXE << 1];
int head[MAXV], tot;
int vis[MAXV], pre[MAXV], path[MAXV], d[MAXV];
int Z[N][N];
int mc, mf;
void init()
{
CLR(head, -1);
tot = 0;
mc = mf = 0;
}
inline void add(int s, int t, int cap, int cost)
{
E[tot] = edge(t, head[s], cap, cost);
head[s] = tot++;
E[tot] = edge(s, head[t], 0, -cost);
head[t] = tot++;
}
int spfa(int s, int t)
{
queue<int>Q;
Q.push(s);
CLR(d, INF);
CLR(vis, 0);
d[s] = 0;
vis[s] = 1;
while (!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = 0;
for (int i = head[u]; ~i; i = E[i].nxt)
{
int v = E[i].to;
if (d[v] > d[u] + E[i].cost && E[i].cap > 0)
{
d[v] = d[u] + E[i].cost;
pre[v] = u;
path[v] = i;
if (!vis[v])
{
vis[v] = 1;
Q.push(v);
}
}
}
}
return d[t] != INF;
}
void MCMF(int s, int t)
{
int i;
while (spfa(s, t))
{
int Min = INF;
for (i = t; i != s; i = pre[i])
Min = min(Min, E[path[i]].cap);
for (i = t; i != s; i = pre[i])
{
E[path[i]].cap -= Min;
E[path[i] ^ 1].cap += Min;
}
mf += Min;
mc += Min * d[t];
}
}
int main(void)
{
int tcase, n, m, i, j, k;
scanf("%d", &tcase);
while (tcase--)
{
init();
scanf("%d%d", &n, &m);
for (i = 1; i <= n; ++i)
for (j = 1; j <= m; ++j)
scanf("%d", &Z[i][j]);
int S = 0, T = n + n * m + 1;
for (i = 1; i <= n; ++i) //n源點到需加工物品
add(S, i, 1, 0);
for (i = n + 1; i <= n + n * m; ++i) //n*m加工時間點到匯點
add(i, T, 1, 0);
for (i = 1; i <= n; ++i)
{
for (j = 1; j <= m; ++j)
{
for (k = 1; k <= n; ++k)//遍歷不同時間點
{
int id = j * n + k;
add(i, id, 1, k * Z[i][j]); //n*m*n
}
}
}
MCMF(S, T);
printf("%.6f\n", mc * 1.0 / n);
}
return 0;
}
POJ 3686 The Windy's（思維+費用流好題）