POJ-3686 The Windy's KM算法 拆点题
参考:https://blog.csdn.net/sr_19930829/article/details/40680053
题意:
有n个订单,m个工厂,第i个订单在第j个工厂生产的时间为t[i][j],同一个工厂可以生产多个订单,但一次只能生产一个订单,也就是说如果先生产a订单,那么b订单要等到a生产完以后再生产,问n个订单用这m个工厂全部生产完需要最少的时间是多少。
思路:
这道题好像用费用流也可以,建图思路好像也是一样的。每个订单耗费时间和在工厂中的等待顺序是有关系的。显然,如果一个工厂有k个订单,那么第一个商品 t1时间,第二个商品就是(t1 + t2)时间,第三个商品就是(t1+t2+t3)...因为我们考虑的是总时间,加起来 = t1 + (t1 + t2) + (t1 + t2 + t3) ... (t1 + t2 ... tk) 。去括号可以发现 K*t1 + (K-1) * t2 + ...tk。但这里你可能还像我一样不知所措。t1 贡献了 K 倍,t2 贡献了(K-1)倍,tk贡献了一倍。说得更清楚一些,某个工厂的倒数第 i 个订单贡献 i * t 的时间。所以我们要给每个工厂开n个点,这个点表示左边某个物品在第(1~n)个时的贡献。就是拆点的思想,每个工厂拆出n种情况。
图片可能更好理解(复制自参考)
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <iomanip> #include <cstdlib> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <cctype> #include <queue> #include <cmath> #include <list> #include <map> #include <set> //#include <unordered_map> //#include <unordered_set> //#include<ext/pb_ds/assoc_container.hpp> //#include<ext/pb_ds/hash_policy.hpp> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l, mid, rt << 1) #define rson (mid + 1, r, rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll> pll; typedef pair<int, int> pii; typedef pair<int, pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q //__gnu_pbds::cc_hash_table<int,int>ret[11]; //这是很快的hash_map #define fi first #define se second //#define endl '\n' #define OKC \ ios::sync_with_stdio(false); \ cin.tie(0) #define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行 #define REP(i, j, k) for (int i = j; i < k; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const double PI = acos(-1.0); template <typename T> inline T read(T &x) { x = 0; int f = 0; char ch = getchar(); while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x = f ? -x : x; } /*-----------------------showtime----------------------*/ const int maxn = 100; int t[maxn][maxn]; int mp[maxn][maxn * maxn]; int visx[maxn], visy[maxn * maxn]; int xn, xm, minz; int linkx[maxn], linky[maxn * maxn]; int wx[maxn], wy[maxn * maxn]; bool dfs(int x) { visx[x] = true; for (int i = 1; i <= xm; i++) { if (!visy[i]) { int t = wx[x] + wy[i] - mp[x][i]; if (t == 0) { visy[i] = true; if (!linky[i] || dfs(linky[i])) { linky[i] = x; linkx[x] = i; return true; } } else if (t > 0) minz = min(minz, t); } } return false; } int km() { for (int i = 1; i <= max(xn, xm); i++) linkx[i] = linky[i] = 0; for (int i = 1; i <= xm; i++) wy[i] = 0; for (int i = 1; i <= xn; i++) { wx[i] = -inf; for (int j = 1; j <= xm; j++) { wx[i] = max(wx[i], mp[i][j]); } } for (int i = 1; i <= xn; i++) { while (true) { memset(visx, 0, sizeof(visx)); memset(visy, 0, sizeof(visy)); minz = inf; if (dfs(i)) break; for (int j = 1; j <= xn; j++) if (visx[j]) wx[j] -= minz; for (int j = 1; j <= xm; j++) if (visy[j]) wy[j] += minz; } } int ans = 0; for (int i = 1; i <= xn; i++) { if (linkx[i] > 0) { ans -= mp[i][linkx[i]]; } } return ans; } int main() { int T; cin >> T; while (T--) { int n, m; scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { scanf("%d", &t[i][j]); } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { for (int k = 1; k <= n; k++) { mp[i][(j - 1) * n + k] = -k * t[i][j]; } } } xn = n, xm = m * n; int ans = km(); printf("%.6f\n", ans * 1.0 / n); } return 0; }
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