hdu4966 GGS-DDU

有 \(n\) 个课程，每种课程有 \(a_i\) 级，一开始你每种课程都为 \(0\) 级，有 \(m\) 个升级方案：\((x,\ l1,\ y,\ l2,\ c)\) ，若你课程 \(x\) 已达到 \(l1\) 级，那么你可以花费 \(c\) 的价格，使得课程 \(y\) 达到 \(l2\) 级。求最小花费使得所有课程满级。

\(n\leq50,\ m\leq2\times10^3,\ a_i\leq500\)

最小树形图

将每个课程拆为 \(a_i\) 个点，分别表示此课程等级为 \(0\cdots a_i\) 。建一个虚拟根节点，连向所有等级为 \(0\) 的节点，边权为 \(0\) 。升级方案可以连边课程 \(x\) 的 \(l1\cdots a_x\) 到课程 \(y\) 的 \(l2\) ，边权为 \(c\) ，再从所有等级为 \(k\) 的节点向等级为 \(k-1\) 的节点连边，权值为 \(0\) ，接着跑最小树形图就吼辣

时间复杂度 \(O(m\sum a_i)\)

代码

#include <bits/stdc++.h>
using namespace std;const int maxn = 2.5e4 + 10, maxm = 1e6 + 10;
int n, m, k, a[60], mp[60][510], val[maxn], vis[maxn], pre[maxn], tid[maxn];
struct edges {int u, v, w;edges(int x = 0, int y = 0, int z = 0) : u(x), v(y), w(z) {}
} e[maxm];int edmonds() {int ans = 0;while (1) {memset(vis, 0, sizeof vis);memset(tid, 0, sizeof tid);memset(val, 0x3f, sizeof val);for (int i = 1; i <= m; i++) {int u = e[i].u, v = e[i].v;if (u != v && e[i].w < val[v]) {val[v] = e[i].w, pre[v] = u;}}for (int i = 1; i < n; i++) {if (val[i] > 1e9) return -1;}int tot = 0;for (int i = 1; i < n; i++) {int u = i;ans += val[i];while (u < n && !tid[u] && vis[u] != i) {vis[u] = i, u = pre[u];}if (u < n && !tid[u]) {tid[u] = ++tot;for (int v = pre[u]; u != v; v = pre[v]) {tid[v] = tot;}}}if (!tot) break;for (int i = 1; i <= n; i++) {if (!tid[i]) tid[i] = ++tot;}for (int i = 1; i <= m; i++) {int u = e[i].u, v = e[i].v;e[i].u = tid[u], e[i].v = tid[v];if (u != v) e[i].w -= val[v];}n = tot;}return ans;
}void solve() {m = 0;for (int i = 1; i <= n; i++) {scanf("%d", a + i);mp[i][0] = mp[i - 1][a[i - 1]] + 1;for (int j = 1; j <= a[i]; j++) {mp[i][j] = mp[i][j - 1] + 1;e[++m] = edges(mp[i][j], mp[i][j - 1], 0);}}for (int i = 1; i <= k; i++) {int p1, p2, l1, l2, w;scanf("%d %d %d %d %d", &p1, &l1, &p2, &l2, &w);for (int j = l1; j <= a[p1]; j++) {e[++m] = edges(mp[p1][j], mp[p2][l2], w);}}for (int i = 1; i <= n; i++) {e[++m] = edges(mp[n][a[n]] + 1, mp[i][0], 0);}n = mp[n][a[n]] + 1;printf("%d\n", edmonds());
}int main() {while (scanf("%d %d", &n, &k) == 2 && n && k) {solve();}return 0;
}